Question

If 0.561 KOH is dissolved in water to give 200 mL, of the solution at 298 K, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH ?

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Solution

Strength space of space solution space equals space fraction numerator 0.561 over denominator 200 end fraction cross times 1000 space gL to the power of negative 1 end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 space cross times space 0.561 space gL to the power of negative 1 end exponent Molar space mass space of space KOH space equals space 39.1 space plus 16 space plus space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 56.1 Molarity space of space solution space equals space fraction numerator 5 space cross times space 0.561 over denominator 56.1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 space cross times space 10 to the power of negative 2 end exponent straight M space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space 0.05 space straight M

KOH being a strong base undergoes almost complete ionisation.

KOH space stack rightwards arrow space with water on top space space straight K to the power of plus left parenthesis aq right parenthesis space plus space OH to the power of minus left parenthesis aq right parenthesis therefore space space space space open square brackets OH to the power of minus close square brackets space equals space open square brackets NaOH close square brackets space equals space 0.05 space straight M space equals space 5 space cross times space 10 to the power of negative 2 end exponent straight M open square brackets straight K to the power of plus close square brackets space equals space 0.05 space straight M Also space space open square brackets straight H to the power of plus close square brackets space open square brackets OH to the power of minus close square brackets space equals space straight K subscript straight w space equals space 1 space cross times space 10 to the power of negative 14 end exponent therefore space space space open square brackets straight H to the power of plus close square brackets space equals space fraction numerator 1 space cross times 10 to the power of negative 14 end exponent over denominator 5 space cross times space 10 to the power of negative 2 end exponent end fraction space equals space 2 space cross times space 10 to the power of negative 13 end exponent straight M Also comma space space pH space equals space minus log space open square brackets straight H to the power of plus close square brackets

equals negative space log space open square brackets 2 space cross times space 10 to the power of negative 13 end exponent close square brackets equals negative open square brackets log space 2 space plus space log space 10 to the power of negative 13 end exponent close square brackets space equals space minus left square bracket 0.3010 space minus space 13 right square bracket space equals space 12.699

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