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

Options

A. B
B. A
C. C
D. D

Answer

C

Explanation

`M_(eq)` of HCl =`75xx1/5xx1=15`
`M_(eq)` of NaOH= `25xx1/5xx1=5`
`M_(eq)` of HCl in resultimg solution =10
Molarity of `[H^(+)]` in resulting mixture
`=(10)/(100)=(1)/(10)`
`pH= -log[H^(+)]= -log[(1)/(10)]=1.0`