What is the pH of a 0.1 M acid solution?

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What is the pH of a 0.1 M acid solution?

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Answer 1 · 2016-06-09T08:23:25

For a strong acid, $p H$ $pH$ $=$ $=$ $- {log}_{10} [H_{3} O^{+}]$ $-log_10[H_3O^+]$ $=$ $=$ $- {log}_{10} [0.1] = - {log}_{10} (10^{- 1})$ $-log_10[0.1]=-log_10(10^-1)$ $=$ $=$ $- (- 1)$ $-(-1)$ $=$ $=$ $1$ $1$

Explanation:

We have assumed a strong acid, for which the equilibrium reaction, as shown below, lies strongly to the right.

$H X (a q) + H_{2} O (l) ⇌ X^{-} + H_{3} O^{+}$ $HX(aq)+H_2O(l)rightleftharpoonsX^(-) + H_3O^(+)$

Why do we use $p H$ $pH$ ? Well it is a holdover from the pre-electronic calculator days, which some of us can remember (not me of course, I am not a day over 29!).

When I write ${log}_{a} b = c$ $log_ab=c$ , I explicitly say that $a^{c} = b$ $a^c=b$ . Common logarithmic bases are $10$ $10$ and $e$ $e$ (I think you cover this in A level mathematics these days). $p H$ $pH$ and $p K_{a}$ $pK_a$ scales use logarithms to the base $10$ $10$ .

Anyway given what I have said ${log}_{10} 100 = 2$ $log_(10)100=2$ , ${log}_{10} 1000 = 3$ $log_(10)1000=3$ , and ${log}_{10} 1 = 0$ $log_(10)1=0$ , and ${log}_{10} 0.1 = {log}_{10} 10^{- 1} = - 1$ $log_(10)0.1=log_(10)10^-1=-1$ . If you can grasp this, you will find the $p H$ $pH$ concept (literally $pouvoir hydrogene, power of hydrogen$ $"pouvoir hydrogene, power of hydrogen"$ ) fairly straightforward.

With weaker acids, say acetic acid, $H_{3} C - C (= O) O H$ $H_3C-C(=O)OH$ , the acid-base equilibrium lies to the left. The equilibrium constant, $K_{a}$ $K_a$ , can be measured and often reported as $p K_{a}$ $pK_a$ $=$ $=$ $- {log}_{10} K_{a}$ $-log_10K_a$ .

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What is the pH of a 0.1 M acid solution?

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