Question

Calculate the pH of each of the following aqueous solutions at 25 °C. (Please help me solve #4 with your work, and check my answers for #1-3, if incorrect please show me why, Thank you)

1) 0.65 M boric acid (B(OH)3, Ka = 7.3 x 10-10) (pH=4.66)

2) 3.15 M ammonia (NH3, Kb = 1.76 x 10-5) (pH= 4.37)

3) 0.82 M benzoic acid (C6H5COOH, Ka = 6.3 x 10-5) (pH=2.14)

4) 0.100 M H3AsO4 (Ka1 = 2.5 x 10-4, Ka2 = 5.6 x 10-8, Ka3 = 3.0 x 10-13) (Please help with this question)

Answer 1

НА + H2О 3 А- + НзО+ A-]H30 К. HA Let [A]H30] [H30+]2 Ка — HA Ka[HA H30*2 -log[K]- log[HA]= -2log[H30*] pKa-log[HA 2pH 1 pHPKloHA = For B(OH)3 we have Ka=7.3 x 10-10 and[HA] = 0.65M. pHpklogHA pHlog7.3x 10-10-log[0.65] H 4.560.09 = 4.653 4.66 For CeH5COOH we have Ka-6.3 x 10-5 and[HA] = 0.82M 1 PH 3D3рКа log[HA pH=log(6.3 x 10-5log0.82 pH = 2.1000.043 2.143

В+ Н2О — ВH+ +он [Bн]Он ] К - В Let [BH+ OH] [ОН ? Къ В] Ks[B] [OH2 log[Kg]-log[B] -2log[OH] pК, — 1og[B] — 2рон 1 РОН 2рк, For NH3 we have KB=1.76 x 105 and[B] = 3.15M. 2PKlogB] РОН log[1.76 x 10-5log3.15 Вля РОН pOH 2.377- 0.249 = 2.128 pH + pОН 3D 14 рH 3D 14 — рОН pH = 11.872

For H3As04 we have Ka1=2.5 x 10-4 and Ka2=5.6 x 10-8 and Ka3=3 x 10-13 [HA] 0.100M Kal >> Ka2>> Ka3 Since Ka1 is much higher than other ka2 and kag the calculation is proceeded by considering the value of Ka1 alone. 1 pH 2PK log[2.5 x 10-4]log[0.1] log [HA pH pH 1.800.5 = 2.3