Question

A solution has a [OH-] = 5.50 x 10-14 M, calculate the pH, [H+], and pOH. Classify the solution as an acid, base, or neutral solution.

Answer 1

Given :

$\sf \: \: [OH^{-}] \: = 5.5 \times \: {10}^{ - 14}$

To find :

• pH
• [H+]
• pOH

Formula used :

$\sf \bullet \: \: pOH = \:- \log [OH^{-}] \\ \\ \sf \bullet \: \: pH =\: - \log [H^+] \\ \\ \sf \bullet \: \: pH + pOH = 14 \\ \\ \sf \bullet \: \: [H^{+}] \times [OH^-] = \: 10^{-14}$

Identity used :

$\sf \bullet \: \: log(a \times b) \: = \: log(a) \: + \: log(b) \\ \\ \sf \bullet \: log( {a}^{b} ) \: = \: b \: \times log(a)$

Solution :

$\sf \bullet \: pOH \: = \: - \log[OH^-] \\ \\ \sf \implies \: pOH \: = \: - \: log(5.5 \times {10}^{ - 14}) \: \\ \\ \sf \implies \: pOH \: = \: - ( \: log(5.5) + log( {10}^{ - 14} ) ) \\ \\ \sf \implies \: pOH \: = - ( log(5.5) + ( - 14)( log(10) \: ) \: ) \\ \\ \sf \implies \: pOH \: = - (0.7403 \: - 14 \times 1) \\ \\ \sf \implies \: pOH \: = - (0.7403 - 14) \\ \\ \sf \implies \: pOH \: = - ( - 13.2597) \\ \\ \sf \implies \: pOH \: = \bold{13.2597}$

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We know that , pH + pOH = 14

$\sf \implies \: pH + \: 13.2597 \: = 14 \\ \\ \sf \implies \: pH \: = \: 14 - 13.2597 \\ \\ \sf \implies \: pH \: = \: \bold{0.7403}$

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We know that ,

$\sf \bullet \: \: [H^{+}] \times [OH^-] = \: 10^{-14} \\ \\ \sf \implies \: [H^{+}] \times (5.5 \times {10}^{ - 14} ) = {10}^{ - 14} \\ \\ \sf \implies \: [H^{+}] = \dfrac{ {10}^{ - 14} }{5.5 \times {10}^{ - 14} } \\ \\ \sf \implies \: [H^{+}] = \dfrac{1}{5.5} \\ \\ \sf \implies \: [H^{+}] = \: \bold{0.18}$

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As , pH = 0.7403 ( that is less than 7) , it is Acidic

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ANSWER :

• pH = 0.7403
• [H+] = 0.18
• pOH = 13.2697
• Acidic solution