Question

a) What is the resulting nitrate ion concentration of the diluted solution if 24.00 mL of a 0.514 M sodium nitrate solution is diluted to a total volume of 350.00 mL? Note that this problem is asking for the concentration of the ion (after the salt dissolves) and not the concentration of the salt. Enter units.

b) What is the resulting ammonium ion concentration of the diluted solution if 50.00 mL of a 0.572 M ammonium carbonate solution is diluted to a total volume of 100.00 mL? Note that this problem is asking for the concentration of the ion (after the salt dissolves) and not the concentration of the salt. Enter units.

Answer 1

a.

Before dilution-------------------------- After dilution

M1 = 0.514M      ------------------- M2 =

V1 = 24ml ------------------------------ V2 = 350ml

      M1V1   =   M2V2

         M2   = M1V1/V2

                 = 0.514*24/350   = 0.03525M

NaNO3(aq) -------------> Na^+ (aq) + NO3^- (aq)

0.03525M---------------------------------- 0.03525M

The concentration of nitrate ion = 0.03525M

b.

Before dilution-------------------------- After dilution

M1 = 0.572M      ------------------- M2 =

V1 = 50ml ------------------------------ V2 = 100ml

      M1V1   =   M2V2

         M2   = M1V1/V2

                 = 0.572*50/100 = 0.286M

(NH4)2CO3(aq) ---------------> 2NH4^+ (aq) + CO3^2- (aq)

0.286M -------------------------   2*0.286M

The concentration fo ammonium ion = 0.572M>>>>answer