(A) 60 mL ${M \over {10}}$ HCl + 40 mL ${M \over {10}}$ NaOH

Mili. moles of HCl = 60 $\times$ ${1 \over {10}}$ = 6

Mili. moles of NaOH = 40 $\times$ ${1 \over {10}}$ = 4

Mili. moles of HCl remaining = 6 - 4 = 2

Total volume will be 60 + 40 = 100 mL

Concentration of [H⁺] = ${{2} \over {100}}$ = 2 $\times$ 10^-2

$\therefore$ pH = 2 - log2 = 1.7

(B) 55 mL ${M \over {10}}$ HCl + 45 mL ${M \over {10}}$ NaOH

Mili. moles of HCl = 55 $\times$ ${1 \over {10}}$ = 5.5

Mili. moles of NaOH = 45 $\times$ ${1 \over {10}}$ = 4.5

Mili. moles of HCl remaining = 5.5 - 4.5 = 1

Concentration of [H⁺] = ${{1} \over {100}}$ = 10^-2

$\therefore$ pH = 2

(C) 75 mL ${M \over {5}}$ HCl + 25 mL ${M \over {5}}$ NaOH

Mili. moles of HCl = 75 $\times$ ${1 \over {5}}$ = 15

Mili. moles of NaOH = 25 $\times$ ${1 \over {5}}$ = 5

Mili. moles of HCl remaining = 15 - 5 = 10

Total volume will be 75 + 25 = 100 mL

Concentration of [H⁺] = ${{10} \over {100}}$ = 10^-1

$\therefore$ pH = 1

(D) 100 mL ${M \over {10}}$ HCl + 100 mL ${M \over {10}}$ NaOH

Mili. moles of HCl = 100 $\times$ ${1 \over {10}}$ = 10

Mili. moles of NaOH = 100 $\times$ ${1 \over {10}}$ = 10

Mili. moles of HCl remaining = 10 - 10 = 0

So, it is neutral solution.

$\therefore$ pH = 7