Question

A 671 mL solution of HBr is titrated with 1.01 M KOH. If it takes 1003 mL of the base solution to reach the equivalence point, what is the pH when only 155 mL of the base has been added to the solution?

Answer 1

1st find the concentration of Her

Balanced chemical equation is:

KOH + HBr ---> KBr + H2O

Here:

M(KOH)=1.01 M

V(KOH)=1003.0 mL

V(HBr)=671.0 mL

According to balanced reaction:

1*number of mol of KOH =1*number of mol of HBr

1*M(KOH)*V(KOH) =1*M(HBr)*V(HBr)

1*1.01*1003.0 = 1*M(HBr)*671.0

M(HBr) = 1.5097 M

Given:

M(HBr) = 1.5097 M

V(HBr) = 671 mL

M(KOH) = 1.01 M

V(KOH) = 155 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 1.5097 M * 671 mL = 1013.0087 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 1.01 M * 155 mL = 156.55 mmol

We have:

mol(HBr) = 1.013*10^3 mmol

mol(KOH) = 1.566*10^2 mmol

1.566*10^2 mmol of both will react

remaining mol of HBr = 8.565*10^2 mmol

Total volume = 826.0 mL

[H+]= mol of acid remaining / volume

[H+] = 8.565*10^2 mmol/826.0 mL

= 1.037 M

use:

pH = -log [H+]

= -log (1.037)

= -0.0157

Answer: -0.0157