Question

36. A typical reaction between an antacid and the hydrochloric acid in gastric juice is   NaHCO3(s) + HCl(aq) ⇌ NaCl(aq) + H2O(l) + CO2(g) Calculate the volume (in liters) of CO2 generated from 0.415 g of NaHCO3 and excess gastric juice at 1.00 atm and 37.00°C. 33. Calculate the pH of a 0.17 M CH3COONa solution. (Ka for acetic acid = 1.8 × 10−5.)

Answer 1

NaHCO3(s) + HCl(aq) ⇌ NaCl(aq) + H2O(l) + CO2(g)

1 moles of NaHCO3 react with HCl to gives 1 mole of CO2

84g of NaHCO3 react with HCl to gives 1 mole of CO2

0.415g of NaHCO3 react with HCl to gives = 1*0.415/84   = 0.00494moles of CO2

PV = nRT

P = 1atm

T   = 37C⁰   = 37 + 273 = 310K

V = nRT/P

    = 0.00494*0.0821*310/1   = 0.126L

33.

PH   = 1/2Pka -1/2logc

Pka = -logKa

        = -log1.8*10^-5

       = 4.75

PH   = 1/2 * 4.75-1/2log0.17

       = 2.375-1/2*-0.7695

     = 2.375 +0.38475   = 2.7597