Question

The pH of 10–⁸M of HCl is ​

Answer 1

$\rm \: HCl\:\rightleftharpoons\: \: \underbrace { \: \: H^+}_{ {10}^{ - 8} \: M } \: +Cl^- \:$

$\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf\sf \hookrightarrow \: concentration \: of \: [ {H}^{ + } ] = {10}^{ - 8} \: M \: \\ \\ \sf \: here \: {10}^{ - 8} < {10}^{ - 5} \\ \sf \: so \: we \: must \: have \: add \: {10}^{ - 7} \: M \\ \: \sf with \: the \: concentration \: of \: [ {H}^{ + } ] \: \\ \\ \sf \: so \: total \: concentration \: of \\ \sf \: [ {H}^{ + } ]= {10}^{ - 8} + {10}^{ - 7} = {1.1} \times 10^{ - 7} \: M \\ \\ \sf \: we \: know \: that \\ \\ \rm \: P^H = - \log[ {H}^{ + } ] \\ \\ = - \log(1.1 \times {10}^{ - 7} ) \\ \\ = 6.95 \\ \\ \therefore \rm \: P^H = 6.95 \end{array}}}}$