b. In a titration, a student prepared a solution of sodium carbonate (Na2COs) by

Question

b. In a titration, a student prepared a solution of sodium carbonate (Na2COs) by dissolving 2.50g of the sodium carbonate in a 250 mL volumetric flask. The student then took 25.0 mL portion of the carbonate solution in an Erlenmeyer flask and titrate with HCl solution of unknown molarity from the burette and found that 22.45 mL of HCl is required to neutralize the 25.00 mL solution of the carbonate. i. Determine the molarity of the 250 mL sodium carbonate solution (3 Marks) 25r-tort.sx11 etterty, ourL.0e que 3.5 ii. If the carbonate and the acid reacts according to the equation NapCOg + HC1 + NaCl + CO2 +H2O Calculate the molarity of the HCl solution (5 Marks) .00Y

Explanation / Answer

Ans :

i) Number of moles of sodium carbonate = 2.50 / 105.99

= 0.0236 mol

Molarity = no. of mol of sodium carbonate / volume of solution in L

= 0.0236 / 0.250

= 0.094 M

ii)

The balanced equation is given as :

Na2CO3 + 2HCl = 2NaCl + H2O + CO2

Using the formula M1 V1/ n1= M2 V2 / n2

where M is molarity and V is volume of the solution

0.094 x 25.0 = M2 x 22.45 / 2

M2 = 0.209 M

So the molarity of the HCl solution will be 0.209 M

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