Question

A 50.0 mL solution of 0.156 M KOH is titrated with 0.312 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCI. 0.00 mL pH = 13.2 6.00 mL pH = 13.1 12.5 mL pH = 12.84 20.0 mL pH = 13.35 24.0 mL pH = 13.32 25.0 mL pH = 13.32 26.0 mL 200 mL o pH = pH = 12:37 30.0 mt p# = 12.40 12.37 30.0 mL pH = 12.40

Answer 1

First calculate volume of HCl required to reach the equivalence point

Consider reaction, HCl + KOH $\rightarrow$ KCl + H₂O

From reaction, 1 mol HCl $\equiv$ 1 mol KOH

We have relation, M acid $\times$ V acid = M base $\times$ V base

$\therefore$ V acid = M base $\times$ V base / M acid

V acid = 0.156 M $\times$ 50.0 ml / 0.312 M = 25.0 ml

Volume of acid required to reach equivalence point = 25.0 ml

pH after addition of 20.0 ml HCl

mmol of HCl = concentration $\times$ volume = 0.312 $\times$ 20.0 = 6.24 mmol

mmol of KOH = 0.156 $\times$ 50.0 = 7.8 mmol

mmol of excess KOH = 7.8 - 6.24 = 1.56 mmol

Volume of mixture at this stage = volume of HCl + volume of KOH = 20.0 + 50.0 = 70.0 ml

[KOH] = No. of moles of KOH / volume of solution in L

[KOH] = 1.56 /1000] / [70.0 /1000]

[KOH] = 0.02228 M

We have , pOH = -log [OH^- ] = - log 0.02228 = 1.67

We have relation, pH + pOH = 14

pH = 14 - pOH = 14 -1.67 = 12.33

pH after addition of 24.0 ml HCl

mmol of HCl = concentration $\times$ volume = 0.312 $\times$ 24.0 = 7.488 mmol

mmol of KOH = 0.156 $\times$ 50.0 = 7.8 mmol

mmol of excess KOH = 7.8 -7.488 = 0312 mmol

Volume of mixture at this stage = volume of HCl + volume of KOH = 24.0 + 50.0 = 74.0 ml

[KOH] = No. of moles of KOH / volume of solution in L

[KOH] =[ 0.312/ 1000] / [74.0 /1000]

[KOH] = 0.00422 M

We have , pOH = -log [OH^- ] = - log 0.00422 =2.37

We have relation, pH + pOH = 14

pH = 14 - pOH = 14 -2.37 = 11.63

pH after addition of 25.0 ml HCl (At equivalence point)

At equivalence point, all HCl is consumed by added KOH, solution contain salt KCl. Salt does not undergo hydrolysis, hence pH of solution will be due to dissociation of water. Therefore, pH will be 7.0

pH after addition of 26.0 ml HCl

After equivalence point, there is excess HCl and pH of solution depends on concentration of HCl.

mmol of HCl = 0.312 $\times$ 26.0 = 8.112 mmol

mmol of KOH = 0.156 $\times$ 50.0 = 7.8 mmol

mmol of excess HCl = 8.112 - 7.8 = 0.312

Volume of solution at this stage = volume of HCl + volume of KOH = 26.0 + 50.0 = 76.0 ml

[HCl] = [ 0.312 /1000] /[ 76.0 /1000] = 0.00411 M

We have, pH = -log [H ⁺] = - log 0.00411 = 2.39

pH after addition of 30.0 ml HCl

mmol of HCl = 0.312 $\times$ 30.0 =9.36 mmol

mmol of KOH = 0.156 $\times$ 50.0 = 7.8 mmol

mmol of excess HCl = 9.36 - 7.8 =1.56

Volume of solution at this stage = volume of HCl + volume of KOH = 30.0 + 50.0 =80.0 ml

[HCl] = [ 1.56 /1000] /[ 80.0/1000] = 0.0195 M

We have, pH = -log [H ⁺] = - log 0.0195 =1.73