Question

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 15.0 mL of KOH.

Express your answer numerically.

pH =

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Part C

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 27.0 mL of HNO3.

Express your answer numerically.

pH =

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Part D

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 29.0 mL of NaOH.

Express your answer numerically.

Answer 1

Total number of millimoles of HBr $= 50.0 \times 0.15 = 7.5$ millimoles
Total number of mollimoles of KOH $= 15.0 \times 0.25 = 3.75$ millimoles.
Thus, HBr is present in excess.
Out of 7.5 millimoles of HBr, 3.75 millimoles of HBr will be neutralised with 3.75 millimoles of KOH and
millimoles of HBr will remain unreacted.
Total volume of the solution mL

$[H^+] = \frac {3.75}{65.0} = 0.0577$ M

Part C
The number of millimoles of ammonia $= 75.0 \times 0.200 = 15$ millimoles
The number of millimoles of nitric acid $= 27.0 \times 0.500 = 13.5$ millimoles.
Thus ammonia is present in excess.
Out of 15 millimoles of ammonia, 13.5 millimoles of ammonia will react with 13.5 millimoles of nitric acid to form 13.5 millimoles of ammonium nitrate.
1.5 millimoles of ammonia will remain unreacted.
Thus a basic buffer solution is obtained containing ammonia and ammonium nitrate.
Total volume mL
$[NH_3] = \frac {1.5}{102}$ M
$[NH_4NO_3] = \frac {13.5}{102}$ M
For ammonia, $pK_b = - log K_b = -log 1.8 \times 10^{-5} = 4.74$
$pOH = pK_b + log \frac {NH_4NO_3}{NH_3}$
$pOH = 4.74 + log \frac {\frac {13.5} {102}}{\frac {1.5}{102}}= 5.69$

Part D
The number of millimoles of acetic acid $= 52.0 \times 0.35 = 18.2$ millimoles.
The number of millimoles of NaOH $= 0.40 \times 29.0 = 11.6$ millimoles
Thus acetic acid is present in excess.
Out of 18.2 millimoles of acetic acid, 11.6 millimoles of acetic acid will react with 11.6 millimoles of NaOH to form 11.6 millimoles of sodium acetate.
millimoles of acetic acid will remain unreacted.
An acidc buffer solution is obtained.
Total volume mL
$[CH_3COONa] = \frac {11.6}{81.0}$ M
$[CH_3COOH] = \frac {6.6}{81.0}$ M.
For acetic acid, $pK_a = -log K_a = -log 1.8 \times 10^{-5} = 4.74$
$pH = pK_a + log \frac {[CH_3COONa]}{[CH_3COOH]}$
$pH = 4.74 + log \frac {\frac {11.6}{81.0}}{\frac {6.6}{81.0}} = 4.98$