Question

.65 mL of .24M NaOH are added to 21.6 mL of buffer calculate pH of buffer prior to base addition, after and pH change.

what if instead of buffer it is 21.6 mL of distilled water. ph before, after and the change. 3 sig figs.

buffer was made from .196M HCHOO .271M NaCHOO. with a given ph of 4.4

Answer 1

Given Buffer pH = 4.4 Chcool] - 0.196 M (Na CHOD] = 0.271 M pH of buffer is given by Henderson's Equation pH = pka + log [Na Choo] CHсобн] 4.4 = pka + log 0.27) - 0.196 4.4 = pka t 0.14 pka = 4.4 -0.14 L pka = 4.26

pH before addition de base - 4.40 pka of acid = 4.26 Naon pH after addition Volume of of 0.65 mL 0.24 M Naol = 0.65m milimoles o Naona Molanity X Volume 0:24 X 0.65 W 0.156 milimolej HCOOH HCOO Nat 21-6 ml x 0.271 M 21:6nLX0.196M = 4.2336 = 5.8536 Naonis added, it react with Hoon 5.8536 +0.156 4.2336– 0.156. (=4.0776) € 6.0096 - 6.0096

pH = pka + log CHCOONa] Cacoon] é 4.26 + log 6.0096 4.0776 = 4.26 + 0.168 _ After addition of Naon) pH = 4.428 m 4.43 ph after addition 6 Ngon - 4.43 change in pH = 4.43 - 4.40/ Chinal - initial) 0.030 volume - pH of of water = 31.6 mL neutral water = 7:00 PH before addition of -Naon Volume of Molarity Noon = 0.65 ml of NaOH = 0.24 M

No. o milimoles do Naon - Molavity x volume = 0.247 x 0.65 m2 - 0.156 milimoles. olarity of NaOH = milimoles of Naoh Total volume 0-156 milimoles (216 +0.65 ) mi 0.007 mol/ litre (NaOH) = 0.007 mol/ litse Com] = 0.007 molliitte pOH = -log COH) = -log(0.007) pOH = 2.15 pH = pH = 14- 3.15 11.85

11.85 pH after addition of Naon change in pH = 11.85-7 = 4.85 change in pH = 4.85