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mash [69]
3 years ago
6

During a titration a student found that 20.0cm3 of sodium carbonate solution

Chemistry
1 answer:
vlada-n [284]3 years ago
4 0

Answer: The concentration of Na_2CO_3 required is 0.349mol/dm^3

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity of HNO_3 = 1

M_1 = molarity of HNO_3 solution = 0.500 mol/dm^3

V_1 = volume of  HNO_3 solution = 27.9cm^3

n_2 = acidity of Na_2CO_3 = 2

M_1 = molarity of Na_2CO_3 solution =?

V_1 = volume of  Na_2CO_3 solution =  20.0cm^3  

Putting in the values we get:

1\times 0.500\times 27.9=2\times M_2\times 20.0

M_2=0.349mol/dm^3

Therefore, concentration of Na_2CO_3 required is 0.349mol/dm^3

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