Question

tarefa
Qual o pH de uma solução 0,0003 M de NaOH? Dado = log 3 - Brainly​

Answer 1

• O pH da solução dada é 10,47.

Primeiramente irei calcular o $\sf pOH$. Sabendo que:

                 $\large\displaystyle\begin{gathered} pOH=-\log \left[OH^-\right]\end{gathered}$

Dada a seguinte reação:

                $\large\displaystyle\begin{gathered} NaOH\rightarrow Na^++OH^- \end{gathered}$

      $\large\displaystyle\begin{gathered} 0,0003\ \frac{mol}{l}\rightarrow 0,0003\ \frac{mol}{l}+0,0003\ \frac{mol}{l}\end{gathered}$

• Temos então que:

$\large\displaystyle\begin{gathered} \left[OH^-\right]= 0,0003\ \frac{mol}{l} \end{gathered}$

$\large\displaystyle\begin{gathered} pOH=-\log \left[0,0003\right]\end{gathered}$

$\large\displaystyle\begin{gathered} pOH=-\left[\log \left(3\cdot 10^{-4}\right)\right]\end{gathered}$

$\large\displaystyle\begin{gathered} pOH=-\log (3) - \log (10^{-4})\end{gathered}$

A questão não deu o valor de log3, então irei adotar 0,47.

$\large\displaystyle\begin{gathered} pOH=-0,47 + 4\cdot \log (10)\end{gathered}$

$\large\displaystyle\begin{gathered} pOH=-0,47 +4\end{gathered}$

$\large\displaystyle\begin{gathered} pOH=3,53\end{gathered}$

Feito isso, iremos calcular o $\sf pH$ da seguinte forma:

$\large\displaystyle\begin{gathered} pH+pOH=14\end{gathered}$

$\large\displaystyle\begin{gathered} pH+3,53=14\end{gathered}$

$\large\displaystyle\begin{gathered} pH=14-3,53\end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \therefore\boxed{\boxed{\green{pH=10,47}}}\ \checkmark \end{gathered} }$

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