Here we’ll go over an example where a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid– Strong Base Mixture Calculations Example 1
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Here we’ll go over an example where a strong acid is mixed with a strong base, and we calculate the pH of the final mixture.
Strong Acid–Strong Base
Mixture CalculationsExample 1
We’re given that 150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
And we’re asked to calculate the pH of the final mixture.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
We’ll point out something important here. Strontium hydroxide is a strong base,
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
22Sr 2 HSr OH O
and its formula, Sr(OH)2, has 2 OH’s in it.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
22Sr 2 HSr OH O
So when we write the balanced dissociation equation for Sr(OH)2.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
22
Sr OH Sr 2OH Dissociation Equation
We see that there are 2 moles of OH- for each mole of Sr(OH)2.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
22SrSr OH 2OH
2
2 mol OH
1 mol Sr OH
A conversion factor
So we can use this as a conversion factor
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
22SrSr OH 2OH
2
2 mol OH
1 mol Sr OH
A conversion factor
We’ll start by calculating the initial moles of OH minus added.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OH0.150L 0.0600mol OH
1L 1molSr(mol OH
OH)initial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
We take 0.200 moles of Sr(OH)2 per L
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
2mol OHmol OH 0.150L 0.0600mol OH
1molSr(OH)
0.200molSr(OH)
1Linitial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Multiply it by the conversion factor 2 mol OH- over 1 mole Sr(OH)2.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
2mol0.200molSr(OH)mol
OH
1molSr(OH)OH 0.150L 0.0600mol OH
1Linitial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
And by 0.150 L. We rounded this to 3 significant figures to save room here. The concentration 0.200 mol/L is only 3 significant figures so the answer to this calculation is limited to 3 significant figures.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.0600mol OH
1L 1molSr(0.150L
OH)initial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
When we multiply all three numbers we get 0.0600 mol. So the initial moles of OH minus added is 0.0600 mol. Notice, this is expressed to 3 significant figures, which is consistent with the given data.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L
1L 1molSr(0.0600mol H
OH)Oinitial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Our next step is to calculate the initial moles of H+ added.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
3
3
0.100mol HNO 1mol H0.350L 0.0350mol H
1L 1mol Hmol
NOHinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
The H+ comes from the strong acid nitric acid, or HNO3. Each HNO3 releases 1 proton, so we take 0.100 mol HNO3 per L
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
3
3
1mol Hmol H 0.350L 0.0350mol H
1mol HNO
0.100mol HNO
1Linitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Times 1 mol of H+ per 1 mol of HNO3
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
3
3
1mo0.100mol HNOmol
l H
1mol HNOH 0.350L 0.0350mol H
1Linitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Times 0.350 L. Again we rounded this to 3 significant figures to save room. The number of significant figures in the answer is limited by the 3 significant figures in 0.100 M.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
3
3
0.100mol HNO 1mol Hmol H 0.0350mol H
1L 1mol H0.350L
NOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
The answer comes out to 0.0350 mol. So the initial moles of H+ added is 0.0350 moles.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
3
3
0.100mol HNO 1mol Hmol H 0.350L
1L 1mol H0.0350mo H
NOlinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Notice that in preserving 3 significant figures, both of these have 4 decimal places
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0. mol OH
1L 1molSr(060
OH)0initial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0. mol H
1L 1mol H03
N5
O0initial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Comparing the initial moles of OH minus with the initial moles of H+, we see that we have more moles of OH minus than of H+, so the OH minus is in excess in this case.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L
1L 1molSr(0.0600mol H
OH)Oinitial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Excess
The excess moles of OH minus
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
0.0600mol OH 0.0Excessmol 350mol H 0.0250moO lH OH
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
Is 0.0600 mol OH minus
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L
1L 1molSr(0.0600mol H
OH)Oinitial
0.0600molExcessmol OH 0.0350mol H 0.0250moH l OHO
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
minus 0.0350 mol H+
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
Excessmol OH 0.0600mol O 0.0350mH 0.0250mol Hol H O
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L
1L 1mol H0.0350mo H
NOlinitial
Which equals 0.0250 mol OH minus
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
Excessmol OH 0.0600mol OH 0.0350m 0.0250mol Hol H O
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Notice the numbers we’re subtracting both have 4 decimal places, so our answer must also have 4 decimal places.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
Excessmol OH 0. mol O0600 035H 0. mol H 0. mo0 l O0 H025
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
4 decimal places
4 decimal places
4 decimal places
When expressed to 4 decimal places, this number has 3 significant figures. The zero’s to the left of the 2 are not significant, but the zero after the 5 is.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0 mol0 OH25
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
3 significant figures
Since the hydroxide ion is in excess, we calculate its concentration in the final mixture. It is equal to
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
0.0250mol 0.0250mol
0.0500M0.150L 0.350L 0.500L
OH
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
0.0250 moles
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
Excessmol OH 0.0600mol OH 0.0350m 0.0250mol Hol H O
0.0250mol
OH 0.0500M0.150L 0.350L 0.500
0.0250mol
L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Divided by the total volume of the solution, which is 0.150 L of Strontium hydroxide solution,
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.0600mol OH
1L 1molSr(0.150L
OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.350L 0.500L0.150L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Plus 0.350 L of HNO3 solution.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.350L
0.0250mol 0.0250molOH 0.0500M
0.150L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.0350mol H
1L 1mol H0.350L
NOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
So the concentration of OH minus is 0.0250 moles over 0.500 L
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol
OH 0.0500M0.150L 0.35
0.0250mol
0.5000L L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Which Equals 0.0500 molar
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
0.150L 0.350L 0.500OH 0.0500M
L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Because we have base in excess, we can calculate the pOH
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
log OH log 0.0500 1.30pOH 1 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Which is the negative log of the hydroxide ion concentration
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log 0.0500 1.3log OH 01 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Or the negative log of 0.0500
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
0.150L 0.350L 0.500OH 0.0500M
L
pOH log OH 1.3log 0.05 0100 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Which comes out to 1.301 . None of our data or calculations in this problem have less than 3 significant figures, so the pOH has 3 significant figures or three decimal places.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
1.30pOH log OH log 0. 500 10 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Now we can calculate the pH
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 14.000 pOH 14.000 1.301p 12.699H
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Which is 14 minus the pOH
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.00014.000 1.3p 01 12.H 9O 69
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Or 14 minus 1.301
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
1.30pOH log OH log 0. 500 10 14.000pH 14. 1.3000 pO 01H 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
Which is 12.699. This must has 3 significant figures so its expressed to 3 decimal places.
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
We have now answered the question we set out to answer. The pH of the final mixture is 12.699. This is relatively highly basic which is consistent with the fact that OH- is in excess.
2
2
0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH
1L 1molSr(OH)initial
0.0250mol 0.0250mol
OH 0.0500M0.150L 0.350L 0.500L
pOH log OH log 0.0500 1.301 14.000 pOH 14.000 1.301pH 12.699
3
3
0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H
1L 1mol HNOinitial
Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH
pH of final mixture
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.
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