First slide

PH scale, PH of strong acids and strong bases

Question

One litre of water contains 10^-7 moles of H⁺ ions. Degree of ionisation of water (in percentage) is

Moderate

A

$\large 1.8\, \times {10^{ - 7}}$
B

$1.8\, \times {10^{ - 9}}$
C

$3.6\, \times {10^{ - 7}}$
D

$3.6\, \times {10^{ - 9}}$

Solution

[H⁺] = 10^-7M = C₀α

$\large {H_2}O\left( \ell \right) \rightleftharpoons {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right)$
t=t_eq C₀ - C₀α C₀α C₀α
$\large {K_i} = \frac{{{C_0}{\alpha ^2}}}{{1 - \alpha }} \sim{C_0}{\alpha ^2}\left[ {\alpha < < < 1} \right]$
$\large {\alpha} = \sqrt{\frac{K_i}{C_0}}$
K_i (dissociation constant of water) can be expressed as
$\large {H_2}O\left( \ell \right) \rightleftharpoons {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right)$
$\large {K_i} = \frac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}} = \frac{{{K_w}}}{{55.55M}}$
$\large {K_i} = \frac{{{{10}^{ - 14}}{M^2}}}{{55.55M}} = 1.8 \times {10^{ - 16}}M$
Concentration of pure water at 25^oC is constant and equal to 55.55 M.
K_i = 1.8 $\times$ 10^-16
$\large \alpha = \sqrt {\frac{{1.8 \times {{10}^{ - 16}}}}{{55.55}}} = 0.18 \times {10^{ - 8}} = 1.8 \times {10^{ - 9}}$

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