How do I calculate the $pH$ $"pH"$ of the buffer solution when $1.00$ $1.00$ $g$ $"g"$ of potassium ethanoate is dissolved in $50.0$ $50.0$ ${cm}^{3}$ $"cm"^3$ of $0.20$ $0.20$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ ethanoic acid ( $K_{a} = 1.74 \times 10^{- 5}$ $"K"_"a"=1.74xx10^-5$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ )?

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Answer 1 · 2017-04-19T15:38:25

$pH = 4.77$ $"pH"=4.77$

$ρ$ $rho$ (potassium ethanoate) $= \frac{1}{50 \div 1000} = 20$ $=1/(50divide1000) =20$ $g$ $"g"$ ${dm}^{- 3}$ $"dm"^-3$

$rho="c"xx"Mr"therefore$

[potassium ethanoate] $=20/(39.1+24+32+3)=0.20387$

$["H"^+]="K"_"a"xx[["HA"]]/[["A"^-]]=1.74xx10^-5xx0.2/0.20387=1.7xx10^-5$

$"pH"=-log(1.7xx10^-5)=4.77$