In a titration, 37.65 mL of 0.150 M Na2SO4 solution are needed to completely react with 30.00 mL of a solution

Answers along with the complete solutions would be greatly appreciated, thanks.

2AgNO3 (aq) + Na2SO4 (aq)--->Ag2(SO4) (s) + 2Na(NO3) (aq)

Refer to the equation above to answer the following questions. In a titration, 37.65 mL of 0.150 M Na2SO4 solution are needed to completely react with 30.00 mL of a solution of AgNO3.

A.) How many moles of Na2SO4 were reacted?

B.) How many moles of AgNO3 are in the solution?

C.) What is the molarity of AgNO3 in solution?

Hi Jrm912,


I am french (Boulogne sur mer 62200 - FRANCE)


2AgNO3 (aq) + Na2SO4 (aq)--->Ag2(SO4) (s) + 2Na(NO3) (aq)



A.) How many moles of Na2SO4 were reacted ?
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n(Na2SO4 ) = C x V

n(Na2SO4 ) = 0,150 x 37,65•10^-3

n(Na2SO4 ) = 5,648•10^-3 mole of Na2SO4



B.) How many moles of AgNO3 are in the solution ?
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. . . according to the equation :

1 mole of Na2SO4 requires 2 moles of AgNO3

. . . then :

n(AgNO3) = 2 n(Na2SO4) = 2 x 5,648•10^-3 = 1,130•10^-2 mole of AgNO3



C.) What is the molarity of AgNO3 in solution ?
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n(AgNO3) = C(AgNO3) x V(AgNO3)

. . . then :


C(AgNO3) = n(AgNO3) / V(AgNO3)

C(AgNO3) = 1,130•10^-2 / 30,00•10^-3

C(AgNO3) = 0,3765 mole/L




I hope to have answered your question.
.