Question

A beaker with 1.80×10² mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.30 mL of a 0.310 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Heres what i have so far:

a Hw pH =pka+log / coniugate base L weak acid 5.000 = 41.740 + loul log cab) = S.O60-4.740 ologada 8.26 (b) = 1.8147 [CH₃COOH] + [CH₂ cod) = 0.100M -0.100- 0.06459 x = 0.100-4 x = 0.03546 M. x= 1.8197. Colod-u) 2.3197*4 -0.181477 u = 0.19197 - G.G G454 M 2.8197 5.30mL & 0.001m x 0.310mols H₃O+ = 0.001643 SCH-cool] 1.80x10² mL x 0,00lmL & G 06 454mail = 0,011617 2roles 1.80 x 102ml 0.001 * 0.03546 mol = 0.0663828 maler [CH3COOH) =0.0116172 -0.001643 = 0.0099742 ACH3COOH = 0,6063928-6.001643 4.2 = 0.0047398 V Lotal = 1.804103 +5.30mL = 185.3 mL = 185.3 x 103 [CH₃ CoolH] = 0.004739 / 185.3x103 = 0.025579 [cHacoo] = .0099742 / 185.3x10 - 0.053827 4.740+ (log ( 259)

Answer 1

Г сиз Соо ид wai Cb) 0.03546 1.8 X oX 11/m Maus CCM2 Coou 0.0063828 malus a.gl16i432 nt addud Males O 310 X 5-30 ml mu Oaalc 3 malis pradea (ch loo) to with feait Сиз сов O.001643 hals 0.061k43 0.0G143 males Total No Ca Coon 0.0063818 0.co8025 8 Mals O.001643 + Total valum 185.3 ml + S 3 180 O.1853 L [ сИзсоои) - o.0080258 O.1853L O.o433 M 0.0116172- 0.001643 mals Cb) 0.009974 С сиз соо) . O009974 0-1853

os bban Weak aid Pka t 4.740 t 1 O.0433 00943 .8343 4 83