Question

You added 20 mL of 0.20M solution of Ba(OH)2(aq) to 50 mL of 0.10M solution of HCl(aq). The pH of the resulting solution is ________.

Answer 1

We can solve given problem in following steps.

Step 1 : Calculation of mmol of OH ^- ions and mmol of H ⁺ ions.

mmol of  Ba(OH)₂ = Concentration Volume = 0.20 M 20 ml = 4 mmol

One molecule of Ba(OH)₂ contain two OH ^- ions.

Therefore, mmol of OH ^- ions = 2 ( mmol of  Ba(OH)₂ ) = 8 mmol

mmol of H ⁺ ions = 0.10 M 50 ml = 5.0 mmol

Step 2 : Determination of excess reactant and it's excess concentration.

Consider reaction, H ⁺ + OH ^- H₂O  

According reaction, 1 mmol H ⁺ reacts with 1 mmol OH ^- .  

5.0 mmol H ⁺ reacts with 5 mmol OH ^- .  

Hence, excess mmol of  OH ^- = 8.0 - 5.0 = 3.0 mmol

Volume of solution = 20 ml + 50 ml = 70 ml

[ OH ^- ] = 3.0 mmol / 70 ml = 0.0428 M

Step 3: Calculation of pH

We have relation, pOH = - log [ OH ^- ]

pOH= - log 0.0428

pOH = 1.41

We have relation, pH = 14 - pOH

pH = 14 -1.41

pH = 12.6

ANSWER : pH of solution : 12.6