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drek231 [11]
3 years ago
13

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

ed water. He plans to titrate the acid with solution. A.Calculate the volume of solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
galina1969 [7]3 years ago
6 0

Answer: 36.2 ml

Explanation:

Assmed values :

weight of sulphurous acid = 0.104 g

volume of flask = 250 ml

Molarity of NaOH solution = 0.0700 M

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute  = \frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.104g}{82.07g/mol}=0.00127mole  

V_s = volume of solution in ml

Molarity=\frac{0.00217\times 1000}{250ml}=0.00507M

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of H_2SO_3 solution = 0.00507 M

V_1 = volume of H_2SO_3 solution = 250 ml

M_2 = molarity of NaOH solution = 0.0700 M

V_2 = volume of NaOH solution = ?

n_1 = valency of H_2SO_3 = 2

n_2 = valency of NaOH = 1

2\times 0.00507M\times 250=1\times 0.0700\times V_2

V_2=36.2ml

Therefore, the volume of solution the student will need to add to reach the final equivalence point is 36.2 ml

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