Calculate the pH of the following 2 scenarios a) the change in pH that occurs when...

Calculate the pH of the following 2 scenarios

a) the change in pH that occurs when 1.00g of NaOH is added to 250mL of 0.200M CH₃CO₂H + 0.200M CH₃CO₂Na (answer is changepH=0.477)

b) the pH of the buffer prepared by mixing 2.00g of NaOH with 25.0mL of 3.00M chloroacetic acid when it is diluted to 500mL of water (answer is pH=3.166 but I need to know how to get it)

Solutions

Expert Solution

moles of acetic acid = 250 x 0.2 / 1000 = 0.05

moles of CH3CO2Na = 250 x 0.2 / 1000 = 0.05

moles of NaOH = 1 / 40 = 0.025

pH = pKa + log [salt / acid]

= 4.74 + log [0.05 / 0.05]

pH = 4.74

if 0.025 moles NaOH is added :

pH = pKa + log [salt + C / acid - C]

= 4.74 + log [0.05 + 0.025 / 0.05 - 0.025]

= 5.217

pH change = 5.217 - 4.74 = 0.477

pH change = 0.477

pKa = 2.85

moles of acid = 25 x 3 / 1000 = 0.075

moles of NaOH = 2 / 40 = 0.05

acid + NaOH ----------------> salt + H2O

0.075 0.05 0 0

0.025 0 0.05

pH = pKa + log [salt / acid]

= 2.85 + log [0.05 / 0.025]

pH = 3.16

queen_honey_blossom answered 1 year ago

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