How do you calculate the pH of a weak acid?

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How do you calculate the pH of a weak acid?

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Answer 1 · 2015-09-17T17:08:56

A handy expression to use is $p H = \frac{1}{2} (p K_{a} - log a)$ $pH=1/2(pK_a-loga)$

Explanation:

This is a quick and useful method if you are given the $p K_{a}$ $pK_a$ and concentration $a$ $a$ of the weak acid.

I'll derive it using the usual ICE method:

$H X_{(a q)} ⇌ H_{(a q)}^{+} + X_{(a q)}^{-}$ $HX_((aq))rightleftharpoonsH_((aq))^(+)+X_((aq))^(-)$

$Initial:$ $"Initial:"$ $" "$ a $00$ $" "" "0" " " " " "0$

If $x$ $x$ moles dissociate the equilibrium moles $\Rightarrow$ $rArr$

$(a - x) x x$ $(a-x) " " " " " " " " x " " " " " " x$

$K_{a} = \frac{x^{2}}{a - x}$ $K_a=(x^2)/(a-x)$

We assume that $x$ $x$ is negligible compared to to $a$ $a$ so this approximates to:

$K_{a} = \frac{x^{2}}{a}$ $K_a=(x^2)/(a)$

$\frac{x^{2}}{a} = K_{a}$ $(x^2)/(a)=K_a$

$x^{2} = K_{a} \times a$ $x^2=K_axxa$

$x = {(K_{a} \times a)}_{a}^{\frac{1}{2}}$ $x=(K_axxa)^((1)/(2))$

Taking -ve logs of both sides $\Rightarrow$ $rArr$

$- log x = - \frac{1}{2} log (K_{a} + a)$ $-logx=-1/2log(K_a+a)$

$p H = \frac{1}{2} (p K_{a} - log a)$ $pH=1/2(pK_a-loga)$

Here's an example:

$p H$ $pH$ of a $0.1 M$ $0.1"M"$ solution of ethanoic acid whose $p K_{a} = 4.75$ $pK_a=4.75$ ?

$p H = \frac{1}{2} (4.75 - (- 1))$ $pH=1/2(4.75-(-1))$

$p H = 5.75$ $pH=5.75$

If you can remember it then fine. If not you need to use the ICE method described in the answer by @Dr Hayek

Answer 2 · 2015-09-18T09:28:17

$p H = - log [H_{3} O^{+}]$ $pH = -log[H_3O^+]$

Explanation:

By measuring the concentration of $H_{3} O^{+}$ $H_3O^+$ ions we get the pH using $p H = - log [H_{3} O^{+}]$ $pH = -log[H_3O^+]$

Consider the dissociation of the following weak acid HA with initial concentration of $1.0 M$ $1.0 M$ and $K_{a} = 2.0 \times 10^{- 6}$ $K_a = 2.0 xx 10^-6$ :

$" " " " " " " " " " "$ $H A (a q) + H_{2} O (a q) \to H_{3} O^{+} (a q) + A^{-} (a q)$ $HA(aq) + H_2O(aq) -> H_3O^+(aq) + A^(-)(aq)$
$" " " " " "$ Initial: $1.0 M 0 M 0 M$ $1.0 M " " " " " " " " " " " " 0 M " " " " " " 0M$
$" " " " "$ Change: $- x M +x M + x M$ $- xM " " " " " " " " " " " +x M" " " " " +x M$
$" " "$ Equilibrium: $(1.0 - x) M x M x M$ $(1.0- x)M " " " " " " " " "x M" " " " " " " " x M$

The expression of $K_{a}$ $K_a$ can be written as follows:

$K_{a} = \frac{[H_{3} O^{+} (a q)] \cdot [A^{-} (a q)]}{[H A (a q)]} = \frac{x \cdot x}{1.0 - x} = 2.0 \times 10^{- 6}$ $K_a = ([H_3O^+(aq)]*[A^(-)(aq)])/([HA(aq)])=(x*x)/(1.0-x)=2.0xx10^-6$

solve for $x$ $x$ using calculator, you get

$x \approx 1.4 \times 10^{- 3} \Rightarrow p H = - log (1.4 \times 10^{- 3}) = 2.895$ $x~~1.4xx10^-3 => pH = -log(1.4xx10^-3) = 2.895$

that can be rounded to $p H = 2.9$ $pH=2.9$ (2 significant figures).

Here is a video which summarizes this technique which requires use of the quadratic formula. The values are different than Dr. Hayek's example, but the technique is the same.

Video from: Noel Pauller

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How do you calculate the pH of a weak acid?

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