Question

What is the molarity of a sulfuric acid solution if 60.00 mL of H2SO4 is required to neutralize 0.420 g of sodium hydrogen carbonate (84.01 g/mol)?                                   

H2SO4(aq) + 2 NaHCO3(aq) → Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)

Answer 1

Molar mass of NaHCO3 = 84.01 g/mol

mass of NaHCO3 = 0.42 g

we have below equation to be used:

number of mol of NaHCO3,

n = mass of NaHCO3/molar mass of NaHCO3

=(0.42 g)/(84.008 g/mol)

= 5*10^-3 mol

From balanced chemical reaction, we see that

when 2 mol of NaHCO3 reacts, 1 mol of H2SO4 reacts

mol of H2SO4 reacted = (1/2)* moles of NaHCO3

= (1/2)*5*10^-3

= 2.5*10^-3 mol

This is number of moles of H2SO4

volume , V = 60.0 mL

= 6*10^-2 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2.5*10^-3/6*10^-2

= 4.166*10^-2 M

Answer: 4.166*10^-2 M