Question

A 0.660 gram sample of an unknown monoprotic acid is dissolved in 30.0 mL of water and titrated with a a 0.370 M aqueous potassium hydroxide solution. It is observed that after 10.1 milliliters of potassium hydroxide have been added, the pH is 7.732 and that an additional 5.10 mL of the potassium hydroxide solution is required to reach the equivalence point.

(1) What is the molecular weight of the acid? _______ g/mol

(2) What is the value of K_a for the acid?

Answer 1

Let the Concentration of weak Menoprotic acid be c m Ab equivalente print, weak Menof whic acid CHAI milling of acid - millier of Base (KOMP 30X C = (10.1455.11% 0,27 So 0,187 M) Rencentration of weak acid HA Vkon aloilme TA + oH! la limiting reactant = A 10.1*0,37 + t Ho H. - st 3080-187 وه (Kalha too 4001 4001 he 5.61 ? 40,1 3,737 40.1 Dissociation constant of Ab tak 1,873 40, water) 3.737 니이 V - 707 het hy 1.892 -* * 1,873 tu 83933 tone t 3.737 3.737 - 40,1 a 1,873 40.1 ~ 3.737 401 pH = 7,732 рои - 14 - pң 3/4-7,732 pol a 6.266 no Cole] = lopok 10 6:268 = 5.3958107 ng ax 5:395X/os mx 1,173 Yout

Ka = 3.737 X 107 5.395X/187] Ka 2 0.369x107 [ka = 3069X10587 . Also, [HA]/[weak Monoprotic acid) - 0.187 M granes of acid Molar mass of acid x/2 je oorld7 ler ma Molar mers of acid 0.660 g/mote 0.03 0.187 Pas g/mole 117.65 Molar mans of acid