What is the pH of a 0.0067 M KOH solution?

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Answer 1 · 2016-06-03T02:22:37

$p H = 11.83$ $pH=11.83$

Potassium hydroxide fully ionizes when dissolved in water according to the following equation:

$K O H \to K^{+} + O H^{-}$ $KOH -> K^+ + OH^-$

Now, let's find the relationship between $K O H$ $KOH$ and $O H^{-}$ $OH^-$

$1 : 1 ratio \Rightarrow [K O H] = [O H^{-}] = 6.7 \times 10^{- 3} M$ $" " 1 " : "1 " ratio "=> [KOH] =[OH^-]= 6.7xx10^-3M$

In any aqueous solution, $[H_{3} O^{+}]$ $[H_3O^+]$ and $[O H^{-}]$ $[OH^-]$ must satisfy the following condition:

$[H_{3} O^{+}] [O H^{-}] = K_{w}$ $[H_3O^+] [OH^-]= K_w$

$[H_{3} O^{+}] = \frac{K_{w}}{[O H^{-}]}$ $[H_3O^+] = K_w/([OH^-])$

$[H_{3} O^{+}] = \frac{1.0 \times 10^{- 14}}{6.7 \times 10^{- 3}}$ $[H_3O^+] = (1.0xx10^-14)/(6.7xx10^-3)$

$[H_{3} O^{+}] = 1.5 \times 10^{- 12} M$ $[H_3O^+] = 1.5xx10^-12 M$

Now, after finding the concentration of the hydronium ion, the pH of the solution is determined:

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

$p H = - log [1.5 \times 10^{- 12}]$ $pH = - log[1.5xx10^-12]$

$p H = 11.83$ $pH=11.83$

Other Method
Find the pOH using the concentration of the hydroxide ion, then use the formula $p H + P O H = 14$ $" "pH + POH = 14$ to find the pH.