Question
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/10 KOH, 400 ml M/20 Ba(OH)2 and 200 ml M/20 NaOH if 1100 ml extra water is added to solution.
Solution
The pOH for an aqueous solution is defined in the same way for OH−. The pOH is the negative log of the concentration of hydroxide ions. And pH+pOH=14
Complete step by step answer:
Therefore, moles of OH− in:
300 ml of M/10 KOH =1×300mL×1/10mol/L=30
400 ml 0f M/10 Ba(OH)2=2×400mL×1/20mol/L=40
200 ml 0f M/10 NaOH =200mL×1/20mol/L=10
Volume of extra water =1100mL
Total volume =2000Ml=2L
⇒ [OH−]=200030+40+10=1/25=0.4
⇒ pOH=log[OH−]=−log[1/25]=1.39
⇒ pH+pOH=14
⇒ 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 (H3O+) or hydroxide ions (OH−). 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.