Calculate the pH of a solution obtained by adding $ 5.00\; g$ of pure acetic acid to $ 95.00\; g$ of water ($\rho_{solution} = 1.0055\; \frac{g}{mL}) $ ! Volume of the solution?
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$m_{solution}=100.00\;g$ $V_{solution}=\frac{m_{solution}}{\rho{solution}}=99.45\;mL$
Calculate its initial molarity !
$c_{CH_3COOH}=\frac{5.00}{60\cdot 0.09945}=0.838\;M$
Equation ?
With $x=[H_3O^+]$: $x^2\;+\;10^{-4.75}\;x\;-\;10^{-4.75}\;0.838\;=\;0$ $x\;=\;3.86\cdot 10^{-3} $
pH ?
$pH\;=\;-log\; x\;=\;2.41$
