Question

Find the pH of each of the following solutions of mixtures of acids.

Part A

0.115 M in HBr and 0.145 M in HCHO2

Express your answer to three decimal places.

Part B

0.145 M in HNO2 and 9.0×10⁻² M in HNO3

Express your answer to two decimal places.

Part C

0.190 M in HCHO2 and 0.220 M in HC2H3O2

Express your answer to two decimal places.

Part D

5.5×10⁻² M in acetic acid and 5.5×10⁻² M in hydrocyanic acid

Express your answer to two decimal places.

Answer 1

tal (Hot) = 5.85*103 + 5.85x103 + 6.342 x 10 son 9.006484 pit 6.484 x to DH = -log [ho] = -log[6.484 X10 3 -2.19 Part Di- HC2H2O2(aq) + H20 (1) a Cz hzo (q) + Hot Call 5.58 10 -2 O 1 +2 ta X E 5.58102 hoo Pronla groguen ka [HGH₂2] [C2 H₂02] [hot] na 18x10 = (5.5x102-2) (1.8x105) (5.5x102] X X = 9.98107 010 x 100% 0.000995 9.95810-4 pH = -log [hot] x 100 / [HCher] 9.95X1094 0.055 - = -log (9.95 x10 " 24 3.0

Part A 0. 115M in HBr and 0. 145M in HCHą. the relative strengle of the kião auds. Com mpare Strong acid HBr is HCHR is a weak and So the amount of Hot con contributed by acid is negligible. the weaker + Imol H₂O [H3O+ ] = 0.115 mol HBY Imol Hor IL Sol 0.15 M tell (pH = -log [Hot] - Log [0.115) 0.939 Port Bi- . acid. i mol X so mol HNO3 HNO3 is a strong acid HNO2 is a weak HINO contribution is neglisible. [Hot] - 9.0 xid? mol HNO3 IL Soin 9.00102 =pH = -log[9.0x102) ply --log[us] 11.04 Part Ci- HCHO, and It C2H₂O2 are both weak acids So compare ka valun Hсно, ka value in HC₂ 14₃ OL has a ka value 1.88105 Since the two acids have ka values that differ only wo'. the confrisution of weaker acid Canrot be ignored 1.8810 by So

part club Lag ( + 30t Hсно, (q) + H2011) 2 СНО, 02 o o 2 I 0.190 balawockor ta с -2 x हि 10.190 -2 ) ka = [CHG] [izo costant TEHCHOR] 31 Prone top8 x 10 = [a] [a] (0.190-2) = (1.8x1-4)(0.140962 do a mzurround utores mol HC H₂O2 Laq) + H20 (1) z C2H30 Coq) +118 stuslom w 0.0000342 Verang 0.00585 5. 89 x 10 M 30 + 0.220 5.85*10-3 С E 0.220- 5.85 x 10-3 ka = [(2+₂ Oz] Ligot] [Heltz02] [x][5.85x10 ra] (0.220-2) (1.8x10 | (0.220) = x(5.85 x103 +2] Quadratic solur -4 x= 6.341 x 10 M