Tentukan pH larutan di bawah ini:
(A) 4 gram \(\ce{NaOH}\) dalam 1 liter larutan
(B) \(\ce{Ca(OH)2}\) 0,04 M
Ar Na = 23, O = 16, H = 1
(A) \(\ce{NaOH}\) 0,1 M
Konsentrasi \(\ce{NaOH}\)
\begin{equation*} \begin{split} \text{n} & = \frac {\text{gr}}{\text{Mr}} \\\\ \text{n} & = \frac {4}{23 + 16 + 1} \\\\ \text{n} & = \frac {4}{40} \\\\ \text{n} & = 0,1 \text{ mol} \end{split} \end{equation*}
\begin{equation*} \begin{split} \text{M} & = \frac {\text{n}}{\text{V}} \\\\ \text{M} & = \frac {0,1}{1} \\\\ \text{M} & = 0,1 \end{split} \end{equation*}
Konsentrasi \(\ce{OH-}\)
\begin{array}{ccccc}
\ce{NaOH(aq) & -> & Na+(aq) & + & OH^{-}(aq)} \\
{\color{red}0,1} & & {\color{red} 0,1} & & {\color{red}0,1}
\end{array}
pH larutan
\begin{equation*}
\begin{split}
\text{pOH} & = - \log [\ce{OH-}] \\\\
\text{pOH} & = - \log [0,1] \\\\
\text{pOH} & = - \log \left[10^{-1} \right] \\\\
\text{pOH} & = 1 \\\\
\text{pH} & = 14 - 1 \\\\
\text{pH} & = 13
\end{split}
\end{equation*}
(B) \(\ce{Ca(OH)2}\) 0,04 M
Konsentrai \(\ce{OH-}\)
\begin{array}{ccccc}
\ce{Ca(OH)2(aq) & -> & Ca^{2+}(aq) & + & 2 OH-(aq)} \\
{\color{red}0,04} & & {\color{red} 0,04} & & {\color{red}0,08}
\end{array}
pH larutan
\begin{equation*}
\begin{split}
\text{pOH} & = - \log [\ce{OH-}] \\\\
\text{pOH} & = - \log [0,08] \\\\
\text{pOH} & = - \log \left[8 \:.\: 10^{-2} \right] \\\\
\text{pOH} & = 2 - \log 8 \\\\
\text{pH} & = 14 - (2 - \log 8) \\\\
\text{pH} & = 12 + \log 8
\end{split}
\end{equation*}