Following the solutions were prepared by mixing different volumes of NaOH and HCl of different concentration-pH of which one of them will be equal to 1?

Options

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/(10)NaOH`
D. `100mL.M/(10)HCl+100mL.M/(10)NaOH`

Explanation

`75mLM/5HCL=-75xx1/5=-15m` mol HCl
`25mLM/5NaOH=-25xx1/5=-5m` mol NaOH 5 mmol of NaOH reacts with same mmol of HCl. Hence the HCI remain on the solution = 15 - 5 =10 . Total volume of the solution = 75 + 25 = 100 mL. Thus,
`[HCl]=(10)/(100)=0.1(M)`. Thus pH of the solution
`=-log_(10)0.1=1`