Question

We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 44.5 mL of the NaOH solution to reach the equivalence point.

(a) What is the molar mass of HA?

(b) What is the pK_a value of HA(aq)?

(c) What is the pH at the equivalence point?

(d) What is the pH one third along the titration when you still have twice as much HA(aq) as A^-(aq)?

Answer 1

N.B.- If you have any further queries, please feel free to ask in the comments section.

- b) 0.0ll1a5 mol 25.0x103 L 0.445 m molarity of HA = PH = -4096,0") = -loofkac) -) Skae = 10² 10-PH =10-33 2 Kac ka C = 2.30836 00/1 - 0 30 10 29995 44.5 x 1030.250 moles of NaOH used a roles of HA Present = LL) (mol) O. 011/25 mol molar mass of HA = 2.779 0.011125 mol 248. 988764 glmal. - 249 g/mol (1033) (10-133)2 -2.66 = 4.916318 256 x 10-3 0.445 pka = - dogka = – logly. 916318256x103) = 2.30836001 c) K = = 2.03404 24 4 4 x 10-12 ka 4.916 318256x/0-3 0.011125 mol 225.0+ 44.5x10 3 L = 0.160071942 m [OH-] = 1Kyx C 2.034042444 X10/2 - SJ0607805 x 10-7. POH PH - Pka & = 2.007330015 -14 10-14 [A-] = xo.160071942 -- -log[en] = -log5.7060J6805 x 10-7) – 6.243662387 pH = 14 - POH = 14-624366 2387 = 7.956 337613 ~ 7.76 HA) الله 22.01