Question

Worksheet Week 10 Name: 1. During strenuous exercise lactic acid builds up in a muscle tissues. In a 1.00 Maqueous solution 2.94% of lactic acid is ionized. What is the value of its ka? 2. At equilibrium of (H') in a 0.250 M solution of an unknown acid is 4.07 x 10 M. Determine the degree of ionization and the Ka of this acid. 3. The venom of biting ants contains formic acid, HCOOH, Ka = 1.8 x 10-Mat 25°C. Calculate the pH of a 0.060 M Solution of formic acid

Answer 1

acid HAR НА Н1 + 4 initial concentration final concentration c(l-d) a cd ca dissociation constant, ka (2) (ca) c(1-4) because Ka= THI[A). [HQ] very small, nence 1-4 ~ 1 & is very ka (ca) ((d) a ka= caz wewe co concentration 9 acid intially de degue of dissociation 1. c=10om 0.0294 2= 2.94% = 2.94 100 ka- & 2 cs scanned, 1x 10.0994) - 8,6436x10-4

14+7= ca whue Али &: ka — ср. (шошм Ка - Са-) Гн" 1: Кахсус - 14" 1: 1 Г - ін"). ) кохе Orsoe e ka Чету 6-3 - как 2 50 1 , і х0 - 6 = Ka yo. 250 Ка - ka = Сб. 20 x 106 . , 2 x to 5 | с. 626 x 16-5 1 : 250 4 - 1-62 х 10 б•o1628 3. Гн*. Jка к с. 1-8х10** холово Jн*1. 3. 28 , xio -3 Hн - - log [н'] - - top | 3 15 4 xis * ) CS Canged with pH = 2:48 Scanned with CamScanner

| 4 bH: CH |-4. 1.44 Loi H) TH) = 3.63 x 10² ka = ca ² 14+) = ca ax= [47] - 3.63 x 10° 0.480 x- 0.0756 d in not very small, hence we will not neglect a in the formula: ka= ca 2 Ka = 0.480 x 10.0756² 1-0.0756 Ka : 2.9677 x 10-3 Scanned with CamScanner