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.
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
a) What is the resulting nitrate ion concentration of the diluted solution if 24.00 mL of...
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