Question

Find (H3O), [OH-] for a solution of weak acid HA
ka = 2x10^-8, 8 moles of acid are present in 1 L of solution
​

Answer 1

The concentration of hydronium and hydroxide ions is ${h}^{ + } = 6 \times {10}^{ - 12} M$and $16.66 \times {10}^{ - 4} M$respectively.

Given:

In a solution of weak acid HA, ka = 2x10^-8, 8 moles of acid are present in 1 L of solution.

To Find:

Find (H3O)+ and [OH-].

Solution:

To find the concentration of hydronium and hydroxide ions we will follow the following steps:

As we know,

HA is weak acid so, it contains hydronium ions,

And,

According to the question:

$ka = 2 \times {10}^{ - 8}$

ka is dissociation. constant.

And,

Number of moles = 8

Volume = 1L

$conc = \frac{no \: of \: moles}{volume} = \frac{8}{1} = 8M$

Now,

Relations between pH, PKA, and concentration are:

$ph = \: 7 + \frac{1}{2} ( pka + logc) = 7 + 0.5( - logka + log8)$

$ph =7 + 0.5( - log2 - log {10}^{ - 8} + log {2}^{3} ) = 7 + 0.5( - 0.3010 + 8log10 + 3log2) = 7 + 0.5( - 0.3010 + 8 + 3 \times 0.3010)$

$= 11.301$

Also,

$ph = - log {h}^{ + }$

${h}^{ + } = antilog (- ph) = - 11.301$

${h}^{ + } = 6 \times {10}^{ - 12} M$

Now,

The concentration of OH- ion will be:

${oh}^{ - } = \frac{ {10}^{ - 14} }{ {h}^{ + } } = \frac{ {10}^{ - 14} }{6 \times {10}^{ - 12} } = \frac{1\times {10}^{ - 2} }{6} = 16.66 \times {10}^{ - 4} M$

Henceforth, the concentration of hydronium and hydroxide ions is ${h}^{ + } = 6 \times {10}^{ - 12} M$and $16.66 \times {10}^{ - 4} M$respectively.

#SPJ1