Question

The solubility of Sr(OH)₂ at 298 K is 19·23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

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Solution

Molar space mass space of space Sr left parenthesis OH right parenthesis subscript 2 space equals space 88 space plus space 2 space cross times space 17 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 space space space space equals space 122 space straight g space mol to the power of negative 1 end exponent space space space space space space space space space space

Solubility space of space Sr left parenthesis OH right parenthesis subscript 2 space equals space 19.23 space straight g space straight L 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 space space equals space fraction numerator 19.23 over denominator 122 end fraction space mol space straight L 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 space space equals space 0.158 space mol space straight L to the power of negative 1 end exponent

Assuming complete dissociation,

Sr left parenthesis OH right parenthesis subscript 2 space rightwards arrow space space space Sr to the power of 2 plus end exponent space plus space 2 OH space space rightwards arrow space space Sr to the power of 2 plus end exponent space plus space 2 OH to the power of minus therefore space space space space open square brackets Sr to the power of 2 plus end exponent close square brackets space equals space 0.158 space mol space straight L to the power of negative 1 end exponent space space space space space space space space open vertical bar OH to the power of minus close vertical bar space equals 2 space cross times 0.158 space equals space 0.316 space mol space straight L to the power of negative 1 end exponent

Now comma space space space open vertical bar straight H close vertical bar to the power of plus space equals space fraction numerator straight K subscript straight U over denominator open vertical bar OH to the power of minus close vertical bar end fraction space equals space fraction numerator 1 space cross times space 10 to the power of negative 14 end exponent over denominator 0.316 end fraction space space space space space space space space space space space space space space space space space space space equals space 3.164 space cross times space 10 to the power of negative 14 end exponent therefore space space space pH space equals space minus log space open square brackets straight H to the power of plus close square brackets space equals negative log space left parenthesis 3.164 space cross times space 10 to the power of negative 14 end exponent right parenthesis space space space space space space space space space space equals negative log space 3.164 space minus space log space 10 to the power of negative 14 end exponent space space space space space space space space space space equals space minus 0.5 space plus space 14 space log space 10 space space space space space space space space space space equals space minus 0.5 space plus 14 space equals space 13.50

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