Question

Please answer each question step by step and sometimes I can't read the handwriting, please

1) Calculate the Ka2 for oxalic acid (found in rhubarb and spinach) (H2C2O4) if a 0.35 M solution of Na2C2O4 has a pH of 8.87.

2) An aqueous solution of a weak acid, HWk (pKa = 5.25,) has a pH of 4.53. Calculate the molarity of the weak acid.

3) Calculate the pH of a 0.275 M solution of borax. (a detergent additive often used in areas with hard water) (Ka = 5.8 x 10-10)

4) Calculate the molarity of a solution of nicotine, a weak base, if the solution has a pH = 10.72. Kb = 7.4 x 10-7

Answer 1

1) Na C,0, (aq) → 2Na+ (aq) +C,0; (aq) 0,02(aq) is conjugate base of HC,O pH = 8.87 pOH = 14- pH = 14 -8.87 = 5.13 ToHº 1=10-20 = 10-5.13 = 7.4x10-M C202- is very weak base so the relation between K; , [OH-]and concentration of base. [OH-] = \K,x[C,07"] [OH-] **(0,01] _(7.4x10^M) (0.35M) =1.6x10-10 Ką2 of oxalic acid = _ 1.0x10-14 1.6x10-10 =6.3x10-5

2)pH = 4.53 [H"]=10-P# =104.53 = 3.0x10-M acid is very weak acid so the relation between Ka , [H* )and concentration of acid. [++] = VK_x[acid] (acid] - CH) K. (3.0x10-SM) 10-pka (3.0x10--M) II 10-323 =1.6x10-M

3) acid is very weak acid so the relation between Ka, [H* )and concentration of acid. [ht] =VKx[acid] = 15.8x10-10 (0.275 M) = 1.26x10^M pH = -log[H] =- log (1.26x10-M) = 4.90

4) pH = 10.72 pOH =14 –10.72 = 3.28 [OH-] = 10-P03 = 10-3.38 = 5.3x10-M base is very weak base so the relation between K; , [OH-] and concentration of base. [OH-] = v K, x[base] [base] = [OH-] (5.3x104 M) 7.4x10- =0.30M Concentration of nicotine = 0.38 M