Question

What is the pH (to nearest 0.01 pH unit) of a solution prepared by mixing 76.0 mL of 0.0477 M NaOH and 86.0 mL of 0.0182 M Ba(OH)₂?

Answer 1

Solution :-

Lets first calculate the moles of the NaOH and Ba(OH)2 using the concentrations and volume

moles = molarity * volume in liter

moles of NaOH = 0.0477 mol per L * 0.076 L = 0.003625 mol

moles of Ba(OH)2 = 0.0182 mol per L * 0.0860 L = 0.001565 mol

Ba(OH)2 gives 2 OH- therefore

Total moles of OH- = 0.003525 mol + (0.001565 mol * 2) = 0.006756 mol OH-

now lets calculate the molarity of the OH- at the total volume

total volume = 76.0 ml + 86.0 ml = 162 ml = 0.162 L

molarity of the OH- = 0.006756 mol /0.162 L

                           = 0.0417 M

now lets Calculate the pOH

pOH= -log [OH-]

pOH= - log [0.0417]

pOH= 1.38

now lets find pH

pH +pOH = 14

pH= 14 - pOH

pH= 14 - 1.38

pH= 12.62

Therefore the pH of the solution is 12.62