Question

40 mL of 0.30 weak acid, HA, (Ka = 4.2 X 10^-10), is titrated by 0.200M NaOH. calculate the pH after the following volumes of NaOH have been added.

(a) 0 mL

(b) 44.3 mL

(c) At the equivalence point

(d) 75.0 mL

show work PLEASE. question id due by midnight

Answer 1

C-0-30 M CHA] = yome, 0.301 INuor] = 0.20M, une PHA = (40 x 0:3) naron = (0.2 xu) mmol = 12 mmol ka = 4.2 x 15 10 = pka = -log(42x 17'') pika = 9.37675 a) o me of NaOH s means pure weak Acid HA pH = 4 cpka - lage) =(9.37675 - Song (0:30) [PH = 4.95 Ane 5 44.3 ml of NaOH -, nNaOH = 10.20 x 44.3) > 8.86 mmo) HA + NaOH NGA + H₂O initical 8.86 8.86 - fineel 3.14 - buffer PH= pka + leng CNGAI -) 9.37675 + boy su og Tha] pH = 8.83 | Ane 12

c) at eat point 0 HA = VNĐa 0 14 12 = 0.200 v = v=6ome HA + NaOH - NaA +H2O initial 12 12 final - (NaA] = 12 ) 0:12M csalt of weak deid) 40+60 so pile = { 114+ pla tlege) ={ (14+ 9.39675 + log 0-12) PH = 11-23 Ane 12 mmol D) at 75me of NaOH = n noon = (75X0.20) = ismmo) HA + NaOH - NaA +H2O Initical 15 12 12 final 3 Strong base, so it will dominant the solution over salt of weak And U Cruceron] = volts = M = [OH"] ple = 14+ le con7] a 14+ lokomis) [PH = 12:42 Ane