At a temperature where $K_{w} = 1.0 \times 1 0^{- 13}$ , how many $H X^{+}$ (hydronium) ions are present in $1.0 mL$ of pure water? Express as $z \times 1 0^{14}$ and give $z$ to two decimals ( $N_{A} = 6.022 \times 1 0^{23}$ ; $1 0^{- 0.5} = 0.3162$ ).

Numerical answer type, tolerance 0.02.

Answer: 1.9

Pure water:

[H X^{+}] = K_{w} = 1 0^{- 6.5} = 3.162 \times 1 0^{- 7} mol L^{- 1}

. In

1 mL

3.162 \times 1 0^{- 7} \times 1 0^{- 3} = 3.162 \times 1 0^{- 10} mol

, i.e.

\times 6.022 \times 1 0^{23} = 1.904 e + 14

ions, so

z = 1.90