Answer to Question #165661 in Chemistry for Kei

Question

Calculate the Sodium carbonate

content of 1.12g sample requiring

24.8ml of 0.1947N HCL on titration

with phenolphthalein

Answer 1

NaHCO₃ + HCl = NaCl + H₂O + CO₂

n (NaHCO₃) = n (HCl)

Cn = N/V

N (HCl) = N (NaHCO₃) = V x Cn = 0.0248 x 0.1947 = 0.005 g-equivalents

Number of gram equivalents = weight of solute × [Equivalent weight of solute]^-1

Equivalent weight of solute = M/number of equivalents

M (NaHCO₃) = 84.007 g/mol

Number of equivalents (NaHCO₃) = 2

Equivalent weight of solute (NaHCO₃) = 84.007/2 = 42.007

Weight of solute (NaHCO₃) = Number of gram equivalents x Equivalent weight of solute = 0.005 x 42.007 = 0.21 g

%(NaHCO₃) = 0.21/1.12 x 100 = 18.75%