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Question: Calculate the pH of a solution prepared by mixing 300 ml \(M/10\) KOH, 400 ml \(M/20\) \(Ba{\left( {...

Calculate the pH of a solution prepared by mixing 300 ml M/10M/10 KOH, 400 ml M/20M/20 Ba(OH)2Ba{\left( {OH} \right)_2} and 200 ml M/20M/20 NaOH if 1100 ml extra water is added to solution.

Explanation

Solution

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

Complete step by step answer:
Therefore, moles of OHO{H^ - } in:
300 ml of M/10M/10 KOH​ =  1×300mL×1/10mol/L=30 = \;1 \times 300mL \times 1/10mol/L = 30
400 ml 0f M/10M/10 Ba(OH)2  =  2×400mL×1/20mol/L=40Ba{\left( {OH} \right)_2}\; = \;2 \times 400mL \times 1/20mol/L = 40
200 ml 0f M/10M/10 NaOH =  200mL×1/20mol/L=10 = \;200mL \times 1/20mol/L = 10
Volume of extra water =1100= 1100mL
Total volume =2000Ml=2L = 2000Ml = 2L
\Rightarrow [OH]=30+40+102000=1/25=0.4[OH^-] = \dfrac{{30 + 40 + 10}}{{2000}} = 1/25 = 0.4
\Rightarrow pOH=log[OH]=log[1/25]=1.39pOH = log\left[ {O{H^ - }} \right] = - log\left[ {1/25} \right] = 1.39
\Rightarrow pH+pOH=14pH + pOH = 14
\Rightarrow pH=141.39=12.61pH = 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 (H3O+)\left( {{H_3}{O^ + }} \right) or hydroxide ions (OHO{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<7pH < 7 the solution is considered acidic because of the high concentration of hydrogen ions. At a pH>7pH > 7 the solution is considered basic because of the low concentration of hydrogen ions. At pH=7pH = 7, the solution is considered neutral because hydrogen and hydroxide ion concentration are the same.