Question

Calculate the pH of a solution prepared by mixing 300 ml $M/10$ KOH, 400 ml $M/20$ $Ba{\left( {OH} \right)_2}$ and 200 ml $M/20$ NaOH if 1100 ml extra water is added to solution.

Answer 1

Hint: The pOH for an aqueous solution is defined in the same way for $O{H^ - }$. The pOH is the negative log of the concentration of hydroxide ions. And $pH + pOH = 14$

Complete step by step answer:
Therefore, moles of $O{H^ - }$ in:
300 ml of $M/10$ KOH​ $ = \;1 \times 300mL \times 1/10mol/L = 30$
400 ml 0f $M/10$ $Ba{\left( {OH} \right)_2}\; = \;2 \times 400mL \times 1/20mol/L = 40$
200 ml 0f $M/10$ NaOH $ = \;200mL \times 1/20mol/L = 10$
Volume of extra water $= 1100$mL
Total volume $ = 2000Ml = 2L$
$\Rightarrow$ $[OH^-] = \dfrac{{30 + 40 + 10}}{{2000}} = 1/25 = 0.4$
$\Rightarrow$ $pOH = log\left[ {O{H^ - }} \right] = - log\left[ {1/25} \right] = 1.39$
$\Rightarrow$ $pH + pOH = 14$
$\Rightarrow$ $pH = 14-1.39 = 12.61$

Note: The "p" in pH and pOH indicates "-log" The H or the OH in pH and pOH indicate the concentration of either hydronium ion $\left( {{H_3}{O^ + }} \right)$ or hydroxide ions ($O{H^ - }$). So, pH measures hydronium ion concentration, while pOH measures hydroxide ion concentration. Only strong acids and bases completely dissociate in water. As a result, if you are given the concentration of the acid, you also know the concentration of the hydronium ions. At a $pH < 7$ the solution is considered acidic because of the high concentration of hydrogen ions. At a $pH > 7$ the solution is considered basic because of the low concentration of hydrogen ions. At $pH = 7$, the solution is considered neutral because hydrogen and hydroxide ion concentration are the same.