Question

Hi, please can you show me how to work 5a) and 5b) out. The Ka value is 1,8*10^-5. Thank you.

eai 5. A buffer solution was prepared by adding 4.95 g of sodium acetate, NaCH,Co, to 250 cm'of 0.150 M acetic acid, CH COOH. Ka ,s S (a) What is the pH of the buffer? (b) What is the pH of 100 ml of buffer solution if you add 82 mg of NaOH to the solution? 1201

Answer 1

Mass od sodium actate = 4.95 g NaCH_# CO_2 Molar mass of sodium actate = 82.03 g/mol Therefore Moles of sodium acetate = 4.95/82.03 = 0.06 volume of acetic acid = 250 cm^3 = 0.25 con. Of acetic acid = 0.15 M Molar of acetic acid = 0.25 times 0.15 = 0.0375 as K_a = 1.8 times 10^-5 rho K_a = 4.74 using Handerson-hasselbalch pH = pKa = log(Molas of sodium acetate/Moles of acetic acid) = 4.74 + log(0.06/0.0375) = 4.94 Now, mass of NaOH = 82 mg = 0.082 g Molar mass of NaOH = 40 g/mol Now, Moles of NaOH = 0.082/440 = 0.00205