Question

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :
(1) `60 mL (M)/(10) HCl + 40 mL (M)/(10) NaOH`
2. `55 mL (M)/(10) HCl + 45 mL (M)/(10) NaOH`
`75 mL (M)/(5) HCl + 25 mL (M)/(5) NaOH`
(4) `100 mL (M)/(10) HCl + 100 mL (M)/(10) NaOH`
pH of which one of them will be equal to 1 ?
A. `(2)`
B. `(1)`
C. `(4)`
D. `(3)`

Answer 1

Correct Answer - D
(1) 4 mL `(M)/(10)` NaOH will neutralize 40 mL of `(M)/(10)` HCl .
`(M)/(10)` HCl left unneutralised = 20 mL.
Total volume = 100 mL . Dilution = 5 times.
In final solution, `[HCl]=(M)/(50). pH != 1`.
(2) `(M)/(10)`HCl left unneutralised = 10 mL .
Total volume = 100 mL . Dilution = 10 times. In final solution, `[HCl]=(M)/(100)=10^(-2)M`
`:. pH = 2`.
`(3) (M)/(5) ` HCl left unneutralized = 50 mL .
Total volume = 100 mL . Dilution = 2 times.
In final solution, `[HCl]=(M)/(10)=10^(-1)M`. Hence, pH=1.
(4) There is exact neutralisation. Hence, pH = 7.