Question

a solution of hydrochloric acid of unknown concentration was titrated with .21 M NaOH. if a 75 ml sample of the HCl solution required exactly 13ml of the NaOH solution to reach the equivalence point, what was the ph of the HCl solution

Answer 1

pH of HCl solution is 1.44

Step-by-step explanation:

NaOH is a monoprotic base and HCl is a monobasic acid.

Neutralization reaction:
`NaOH+HCl\rightarrow NaCl+H_(2)O`

According to balanced reaction, 1 mol of NaOH neutralizes 1 mol of HCl.

Number of moles of NaOH in 13 mL of 0.21 M NaOH

=
`(0.21)/(1000)* 13mol=0.00273mol`

let's assume concentration of HCl is C (M)

Then, number of moles of HCl in 75 mL of C (M) HCl solution

=
`(75)/(1000)* Cmol=(75C)/(1000)mol`

So, we can write,
`(75C)/(1000)=0.00273`

or,
`C=0.0364`

1 mol of HCl contains 1 mol of
`H^(+)`

So, concentration of
`H^(+)` in 0.0364 M HCl,
`[H^(+)]=0.0364M`

Hence,
`pH=-log[H^(+)]=-log(0.0364)=1.44`