Question

A 625 mL solution of HBr is titrated with 1.51 M KOH. If it takes 728.5 mL of the base solution to reach the equivalence point, what is the pH when only 137 mL of the base has been added to the solution?

I am getting an answer of .1018, but it's wrong

Answer 1

1st find the molarity of HBr

M(KOH)=1.51 M

V(KOH)=728.5 mL

V(HBr)=625.0 mL

According to balanced reaction:

number of mol of KOH =number of mol of HBr

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

1.51*728.5 = M(HBr)*625.0

M(HBr) = 1.76 M

Now find the pH

Given:

M(HBr) = 1.76 M

V(HBr) = 625 mL

M(KOH) = 1.51 M

V(KOH) = 137 mL

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

mol(HBr) = 1.76 M * 625 mL = 1100 mmol

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

mol(KOH) = 1.51 M * 137 mL = 206.87 mmol

We have:

mol(HBr) = 1100 mmol

mol(KOH) = 206.87 mmol

206.87 mmol of both will react

remaining mol of HBr = 893.13 mmol

Total volume = 762.0 mL

[H+]= mol of acid remaining / volume

[H+] = 893.13 mmol/762.0 mL

= 1.1721 M

use:

pH = -log [H+]

= -log (1.1721)

= -0.0690

Answer: -0.0690