Start by determining the equivalence point volume for this titration. A differen

Question

Start by determining the equivalence point volume for this titration. A different method will be required to calculate the pH of the solution depending on whether the added volume is before, after, or at the equivalence point.

The pKa of hypochlorous acid is 7.530. A 50.0 mL solution of 0.108 M sodium hypochlorite (NaOCI) is titrated with 0.339 M HCI. Calculate the pH of the solution a) after the addition of 6.15 mL of 0.339 M HC Number pH- b) after the addition of 16.6 mL of 0.339 M HC Number c) at the equivalence point with 0.339 M HC Number PH-

Explanation / Answer

Sodium hypochlorite; NaOCl is ionizing as following:

NaOCl ó Na+   OCl-

First calculate the Kb for OCl- as follows:

Kb = 10^-6.44 = 3.63 x 10^-7

ClO- + H2O <----> HClO + OH-

Kb =[ HClO +] [ OH-] / [ClO-]

Assume =[ HClO +]= [ OH-] =x
3.63 x 10^-7 = x^2 / 0.108-x
x = [OH-]= 0.000198 M
pOH = 3.70

pOH+ pH=14
pH = 10.3

moles ClO- = 0.050 L x 0.108 M=0.00540

moles H+ = 6.15 x 10^-3 L x 0.339 M=0.00208
ClO- + H+ = HClO
moles ClO- = 0.00540 - 0.00208 = 0.00332
moles HClO = 0.00208
total volume = 0.05615 L

[ClO-]= 0.00332/ 0.05615=0.057 M
[HClO]= 0.00208/ 0.05615=0.037 M
pKa = 7.530

pH = pKa+ log salt / acid
pH = 7.530 + log 0.057/ 0.037=7.72

First calculate the Kb for OCl- as follows:

Kb = 10^-6.44 = 3.63 x 10^-7

ClO- + H2O <----> HClO + OH-

Kb =[ HClO +] [ OH-] / [ClO-]

Assume =[ HClO +]= [ OH-] =x
3.63 x 10^-7 = x^2 / 0.108-x
x = [OH-]= 0.000198 M
pOH = 3.70

pOH+ pH=14
pH = 10.3

moles ClO- = 0.050 L x 0.108 M=0.00540

moles H+ = 16.6 x 10^-3 L x 0.339 M=0.0056
ClO- + H+ = HClO
moles ClO- = 0.0056 - 0.00540= 0.0002
moles HClO = 0.0056
total volume = 0.0666 L

[ClO-]= 0.0002/ 0.05615=0.003 M
[HClO]= 0.0056/ 0.05615=0.084 M
pKa = 7.530

pH = pKa+ log salt / acid
pH = 7.530 + log 0.003/ 0.084=6.08

The equivalent point is defined as the point where the moles of strong acid added = initial moles of base B in solution:

moles ClO- = 0.050 L x 0.108 M=0.00540

moles of HCl is calcaulted as follows:

0.339 mol / 1000 mL *x= 0.00540 mol

x= 0.00540 M *1000 / 0.339 mol=15.9 mL

moles of HCl = 15.9/1000x 0.339 = 0.00540 mol

total volume = 0.0659 L

[ClO-]= 0.00540/ 0.0659=0.082 M
[HClO]= 0.0040/ 0.05659=0.082 M
pKa = 7.530

pH = pKa+ log salt / acid
pH = 7.530 + log 0.082/ 0.082=7.530

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