Question

10mL of 1M silver nitrate is combined with 25mL of 0.1M sodium chromate. What is the concentration of sodium ion after the two solutions are combined? This set up will be used for the next two questions as well, so you should also calculate the combined concentrations of the other ions.

Answer 1

Answer:

Concentration of sodium ions in final solution is 0.1428 mol/L.

Concentration of nitrate ions in final solution is 0.2857 mol/L.

Explanation:

Concentration =$\frac{\text{Moles of compound}}{\text{Volume of the solution (L)}}$

$Na_2CrO_4(aq)+2AgNO_3(aq)\rightarrow Ag_2CrO_4(s)+2NaNO_3(aq)$

Moles of silver nitrate in 10 mL of 1 M silver nitrate.

Volume of the  silver nitrate solution = 10 mL = 0.010 L

$1 M =\frac{\text{Moles of}Ag_2CrO_4}{\text{Volume of the solution in L}}$

Moles of silver nitrate =$1 M\times 0.010 L=0.010 mol$

Moles of sodium chromate in 25 mL of 0.1 M silver nitrate.

Volume of the  silver chromate solution = 25 mL = 0.025 L

$0.1 M =\frac{\text{Moles of}Na_2CrO_4}{\text{Volume of the solution in L}}$

Moles of sodium nitrate =$0.1 M\times 0.025 L=0.0025 mol$

According to reaction, 1 mole of sodium chromate reacts with 2 mole of silver nitrate.

Then , 0.0025 mol of sodium chromate will react with :

$\frac{1}{2}\times 0.0025 mol=0.00125 mol$ of silver nitrate.

1 mol of sodium chromate gives 2 mol of sodium ions and 1 mol of chromate ions.

Then 0.0025 mol of sodium chromate will give 0.0050 mol of sodium ions.

Volume of the solution after mixing =

10 mL + 25 mL = 35 mL =0.035 L

Concentration of sodium ions in final solution:

$[Na^+]=\frac{0.0050 mol}{0.035 mL}=0.1428 mol/L$

Concentration of sodium ions = 0.1428 mol/L

1 mole of silver nitrate gives 1 mol of silver ion and 1 mole nitrate ion

Then 0.010 moles of silver nitrate will give 0.010 moles of nitrate ions.

Concentration of nitrate ion in the final solution:

$[NO_3^{-}]=\frac{0.010 mol}{0.035 L}=0.2857 mol/L$

Concentration of nitrate ions = 0.2857 mol/L

Answer 2

The concentration of sodium ion after the two solutions are combined is 0.1428M.

How we calculate concentration?

Concentration will be calculated in terms of molarity as:

M = n/V, where

n = no. of moles

V = volume

Given chemical reaction is:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

Given volume of AgNO₃ = 10mL = 0.010L

Molarity of AgNO₃ = 1M

Moles of AgNO₃ = 0.010×1 = 0.010 moles

Given volume of Na₂CrO₄ = 25mL = 0.025L

Molarity of Na₂CrO₄ = 0.1M

Moles of Na₂CrO₄ = 0.025×0.1 = 0.0025 moles

From the stoichiometry of the reaction it is clear that:

1 mole of Na₂CrO₄ = produce 2 moles of sodium ions

0.0025 moles of Na₂CrO₄ = produce 2×0.0025=0.0050 moles of sodium ions

Total volume of final solution = 25mL + 10mL = 35mL = 0.035L

Now we calculate the concentration in terms of molarity of sodium ions as:
M = 0.0050/0.035 = 0.1428M

Hence, the concentration of sodium ions is 0.1428M.

To know more about molarity, visit the below link:
https://brainly.com/question/26873446