Lecture 24 10/31/05. Finding endpoint with pH electrode.

Lecture 2410/31/05

Finding endpoint with pH electrode

Titration of H6A with NaOH

Gran Plot

)VV(K10V beaA

HApHb

Advantage is that you can use data before the endpoint to find the endpoint

Vb never goes to 0 because 10-pH never gets to 0

Also slope doesn’t stay constant as Vb nears 0

Indicator• Weak acid or base

• Different protonated forms have different colors• Need a ratio greater than ~ 10:1 to see color

• Seek indicator whose color change is near equivalence point

1pKpH ,1.0]HIn[

]In[ when

1pKpH ,10]HIn[

]In[ when

]HIn[

]In[logpKpH

Indicator errors• Difference between endpoint (color

change) and true equivalence point

• If you use too much can affect reaction

Problem 7-11The carbonate content of 0.5413g of powdered

limestone was measured by suspending the powder in water, adding 10.00 mL of 1.396 M HCl, and heating to dissolve the solid and expel CO2:

CaCO3(s) [FM 100.087] + 2H+ Ca2+ + CO2(g) + H2O

The excess acid required 39.96 mL of 0.1004M NaOH for complete titration to a phenolphthalein end point. Find the weight % of calcite in the limestone.

Problem 7-11 (solutions)Moles OH- = (39.96 mL)*(0.1004 M) = 4.012 mmol

Moles H+ = (10 mL)*(1.396 M) = 13.96 mmol

Moles H+ used to titrate CaCO3 = 9.948 mmol

Moles CaCO3 = 9.948 mmol H*(1 mol CaCO3 / 2 mol H)

Moles CaCO3 = 4.974 mmol

Mass CaCO3 = 4.974 mmol *(100.087 g/mol) = 0.498 g

Weight % = 0.498 g / 0.5413 * 100 = 92%

Quiz 4

A sample was analyzed using the Kjeldahl procedure. The liberated NH3 was collected in 5.00 mL of 0.05 M HCl, and the remaining acid required 3 mL of 0.035 M NaOH for a complete titration. How many moles of Nitrogen were in the original sample?