d
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\)

-or \(\quad\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} ;\left[\mathrm{H}^{+}\right]\) of solution \(1=10^{-3}\)

\(\left[\mathrm{H}^{+}\right]\) of solution \(2=10^{-4} ;\left[\mathrm{H}^{+}\right]\) of solution \(3=10^{-5}\)

Total concentration of \(\left[\mathrm{H}^{+}\right]\)

\(=10^{-3}\left(1+1 \times 10^{-1}+1 \times 10^{-2}\right)\)

\(\Rightarrow 10^{-3}\left(\frac{1}{1}+\frac{1}{10}+\frac{1}{100}\right)\)

\(\Rightarrow 10^{-3}\left(\frac{100+10+1}{100}\right)\)

\(\Longrightarrow 10^{-3}\left(\frac{111}{100}\right)=1.11 \times 10^{-3}\)

\(So, \mathrm{H}^{+}\) ion concentration in mixture of equal volume of these acid solution \(=1.11 \times 10^{-3} / 3=3.7 \times 10^{-4} \,\mathrm{M}\)