Question

Using a 0.20 M phosphate buffer with a pH of 6.4, you add 0.71 mL of 0.54 M HCl to 55 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.) 49 Using a 0.20 M phosphate buffer with a pH of 6.4, you add 0.71 mL of 0.54 M NaOH to 55 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.) Supporting Materials

Answer 1

phosphate butten pH = 6.4 If the concentration of both the same then pl = pka = 6-4 salt and acid is

pH = pka + doy [H poj?) EHPON] Initially the butter contains 55 mL of 0.2m of both upou and H₂ Pou .. 0.055 LX0-2 mol [ of tpol and H₂ por 0.011 mole of both tpol and ltz Pou when we add 0.71 ml of 0.54 Hel to the butten i.e 0.00071L X 0.54 molhl= 3.83x104 mole of thel then HPO + H₂ot - H₂ pou +H2O decreased by increased by - 3.83x10 mol 3.83x10 "mol so Aften addition of Hel THPO, 2-] = (0.011 - 3.83x109) mole = 0.0106 mole [H2 POW] = (0.011 + 3.83X104) mole = 0.01138 mole . i .:. pH = pka + log pH = pka + loy [HPO4?] TH₂ Por]

PH 6.4 + dog (0.0106 ) olo.01138 ) = 6.4 -0.0308 pH = 6.092 PH slightly decreases.

Similarly when add 0.71 ml of 0.54 Much to the butter i.e. o.600712 x 0.54 molt' = 3.83X104 mole of Nach then н, Рок тон — HPo4 + H, 0 decreased by increased by 3.83x10" mole 3.83x10" mole So After addition of Naoh (HPO42-) = (0.011+ 31838164) mole = 0.01138 mole [H₂ poü} = (0.011 - 3.8 3 x104) mole = 0.0106 male :. pH = pka + log " = pka + loy [HPO4?) Hapoi] = 6.4 + dog 10.01138) to. 0106) = 6.4 +0.03059 = 6.4305 pH slightly increases.