Question

The volume of 0.10 M-AgNO$_3$ should be added to 10.0 ml of 0.09 M-K$_2$CrO$_4$ to precipitate all the chromate as Ag$_2$CrO$_4$ is

• A. 18 ml
• B. 27 ml
• C. 9 ml
• D. 36 ml

Answer 1

Answer: (A)

18 ml

Answer 2

The precipitation reaction between silver nitrate ($AgNO_3$) and potassium chromate ($K_2CrO_4$) to form silver chromate ($Ag_2CrO_4$) is: $2Ag^+(aq) + CrO_4^{2-}(aq) \rightarrow Ag_2CrO_4(s)$ From the stoichiometry, 2 moles of $Ag^+$ are required for every 1 mole of $CrO_4^{2-}$.

1. Calculate moles of chromate ions ($CrO_4^{2-}$): The volume of $K_2CrO_4$ solution is 10.0 ml, which is $0.010$ L. The molarity of $K_2CrO_4$ solution is $0.09$ M. Moles of $K_2CrO_4$ = Molarity $\times$ Volume Moles of $K_2CrO_4$ = $0.09 \, \text{mol/L} \times 0.010 \, \text{L} = 0.0009 \, \text{mol}$. Since $K_2CrO_4$ dissociates into $2K^+$ and $CrO_4^{2-}$, the moles of $CrO_4^{2-}$ are equal to the moles of $K_2CrO_4$. Moles of $CrO_4^{2-}$ = $0.0009 \, \text{mol}$.

2. Calculate moles of silver ions ($Ag^+$) required: Based on the stoichiometry ($2:1$ ratio of $Ag^+$ to $CrO_4^{2-}$), the moles of $Ag^+$ required are: Moles of $Ag^+$ = $2 \times \text{Moles of } CrO_4^{2-}$ Moles of $Ag^+$ = $2 \times 0.0009 \, \text{mol} = 0.0018 \, \text{mol}$.

3. Calculate the volume of $AgNO_3$ solution: The molarity of the $AgNO_3$ solution is $0.10$ M. Let $V$ be the volume of $AgNO_3$ solution in liters. Moles of $Ag^+$ from $AgNO_3$ = Molarity of $AgNO_3 \times$ Volume of $AgNO_3$ $0.0018 \, \text{mol} = 0.10 \, \text{mol/L} \times V$ $V = \frac{0.0018 \, \text{mol}}{0.10 \, \text{mol/L}} = 0.018 \, \text{L}$.

4. Convert volume to milliliters: $V = 0.018 \, \text{L} \times 1000 \, \text{ml/L} = 18 \, \text{ml}$.