Question

Calculate the theoretical pH of the solution when you vegin with 10 mL of 0.200 M HCL and dilute it to 50 mL with distilled water. Then calcualte the pH after addition of 5.00, 10.00, 15.00, 20.00 and 25.00 mL of 0.100 M NaOH to the solution above.

Answer 1

First calculate concentration of diluted 50 ml HCl

Use the formula C₁V₁ = C₂V₂

Where C₁ = initial concentration = 0.200 M

V₁ = initial volume = 10 ml

C₂ = final concentration = ?

V₂ = final volume = 50 ml

Substitute the value in formula

C₁V₁ = C₂V₂

C₂ = C₁V₁/V₂

= 0.200 X 10/50

C₂ = 0.04 M

# no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.04 X 0.050 = 0.002 mole

no. of mole of NaOH = 0.100 X 0.005 = 0.0005 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.0005 moel of HCl react with 0.0005 mole of NaOH

no of mole of HCl remain in solution = 0.002 - 0.0005 = 0.0015 mole

total volume of solution = 50 + 5 = 55 ml = 0.055 liter

molarity = no. of mole / volume of solution

molarity of HCl = 0.0015 / 0.055 = 0.027 M

HCl is strong acid dissociate completly therefore [HCl] = [H⁺​]

pH = -log[H⁺] = -log(0.027) = 1.57

pH = 1.57

# no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.04 X 0.050 = 0.002 mole

no. of mole of NaOH = 0.100 X 0.010 = 0.001 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.001moel of HCl react with 0.001 mole of NaOH

no of mole of HCl remain in solution = 0.002 - 0.001 = 0.001 mole

total volume of solution = 50 + 10 = 60 ml = 0.060 liter

molarity = no. of mole / volume of solution

molarity of HCl = 0.001 / 0.060 = 0.016 M

HCl is strong acid dissociate completly therefore [HCl] = [H⁺​]

pH = -log[H⁺] = -log(0.016) = 1.80

pH = 1.80

# no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.04 X 0.050 = 0.002 mole

no. of mole of NaOH = 0.100 X 0.015 = 0.0015 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.0015 moel of HCl react with 0.0015 mole of NaOH

no of mole of HCl remain in solution = 0.002 - 0.0015 = 0.0005 mole

total volume of solution = 50 + 15 = 65 ml = 0.065 liter

molarity = no. of mole / volume of solution

molarity of HCl = 0.0005 / 0.065 = 0.00769 M

HCl is strong acid dissociate completly therefore [HCl] = [H⁺]

pH = -log[H⁺] = -log(0.00769) = 2.11

pH = 2.11

# no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.04 X 0.050 = 0.002 mole

no. of mole of NaOH = 0.100 X 0.02 = 0.002 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.002 moel of HCl react with 0.002 mole of NaOH

complete neutrilization take place and both HCl is strong acid and NaOH is strong base therefore pH = 7

pH = 7

# no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.04 X 0.050 = 0.002 mole

no. of mole of NaOH = 0.100 X 0.025 = 0.0025 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.002 moel of HCl react with 0.002 mole of NaOH

no of mole of NaOH remain in solution = 0.0025 - 0.002 = 0.0005 mole

total volume of solution = 50 + 25 = 75 ml = 0.075 liter

molarity = no. of mole / volume of solution

molarity of NaOH = 0.0005 / 0.075 = 0.00667 M

NaOH is strong base dissociate copletly therefore [NaOH] = [OH^-]

pOH = -log[OH^-] = -log(0.00667) = 2.18

pH = 14 - pOH = 14 - 2.18 = 11.82

pH = 11.82