In: Chemistry
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)
a)
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
b)
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