Predicting a precipitate based on Ksp

Question

$\pu{105 mL}$ of $\pu{0.10 M}$ $\ce{AgNO3}$ is added to $\pu{125 mL}$ of $\pu{0.35 M}$ $\ce{K2CrO4}.$ Will a precipitate form? $(K_\mathrm{sp}$ for $\ce{Ag2CrO4}$ is $\pu{1.12E-12}.)$

Answer:

Yes, $Q = \pu{4.0E-4}$ which is greater than $K_\mathrm{sp}.$

Can someone please help me out with this question? I know that I have to find $Q_\mathrm{sp}$ and compare this to the $K_\mathrm{sp}$ given, and if $Q_\mathrm{sp}$ is bigger, then a precipitate will form. However, I do not know how to find $Q_\mathrm{sp}$ when two substances are mixed together.

Answer 1

To obtain $Q_\mathrm{sp}$ for

$$\ce{Ag2CrO4 <=> 2 Ag+ + CrO4^2-}$$

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CrO4^2-}]$$

$$Q_\mathrm{sp} = c(\ce{Ag+})^2\cdot c(\ce{CrO4^2-})\label{eqn:1}\tag{1}$$

you need to use new concentrations of ions in solution $c(\ce{Ag+})$ and $c(\ce{CrO4^2-})$, which can be found from the corresponding values for initial concentration $c_0$ and volume $V_0$:

$$ \begin{align} c(\ce{Ag+}) &= \frac{c_0(\ce{AgNO3})\cdot V_0(\ce{AgNO3})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\ &= \frac{\pu{0.10 mol L-1} × \pu{105 mL}}{\pu{105 mL} + \pu{125 mL}}\\ &= \pu{4.57E-2 mol L-1} \label{eqn:2}\tag{2} \end{align} $$

$$ \begin{align} c(\ce{CrO4^2-}) &= \frac{c_0(\ce{K2CrO4})\cdot V_0(\ce{K2CrO4})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\ &= \frac{\pu{0.35 mol L-1} × \pu{125 mL}}{\pu{105 mL} + \pu{125 mL}} \\ &= \pu{1.90E-1 mol L-1} \label{eqn:3}\tag{3} \end{align} $$

Finally, plug in the concentrations of ions from \eqref{eqn:2} and \eqref{eqn:3} in \eqref{eqn:1} and you find a value for $Q_\mathrm{sp},$ which indeed exceeds $K_\mathrm{sp}$:

$$Q_\mathrm{sp} = (\pu{4.57E-2 mol L-1})^2 × \pu{1.90E-1 mol L-1} = \pu{4.00E-4 mol3 L-3}$$

Answer 2

You are in right track on this problem. But you admitted that you don't know how to find the concentrations, when two substances (of solutions) are mixed together. First, no other information is given, you have to assume that these two solutions are additive. Then, you can treat it as a dilution.

For example, if $v_1$ volume of a solution with $c_1$ concentration of one substance is added together with $v_2$ volume of a solution with $c_2$ concentration of another substance, we can assume final volume is $v_1+v_2$ (only if volumes are additive). Now, you can treat this as you are diluting $v_1$ volume of a solution with $c_1$ concentration of one substance to $v_1+v_2$ volume. If the final concentration of the substance is $c_f$ after the dilution, you can use $c_iv_i=c_fv_f$ for this calculation.

Now, let's look at the problem in hand: $\pu{105 mL}$ $\left\{v_i(\ce{AgNO3})\right\}$ of $\pu{0.10 M} \; \ce{AgNO3}$ $\left\{c_i(\ce{AgNO3})\right\}$ is added to $\pu{125 mL}$ $\left\{v_i(\ce{K2CrO4})\right\}$ of of $\pu{0.35 M} \; \ce{K2CrO4}$ $\left\{c_i(\ce{K2CrO4})\right\}$. The final volume for both dilution is: $$ v_f = \left\{v_i(\ce{AgNO3})\right\} + \left\{v_i(\ce{K2CrO4})\right\} = \pu{105 mL} + \pu{125 mL} = \pu{230 mL}$$ Thus, we can calculate the concentration of $\ce{AgNO3}$ after dilution using $c_iv_i=c_fv_f$: $$c_f\left\{(\ce{AgNO3})\right\} = \frac{c_i\left\{(\ce{AgNO3})\right\} \times v_i\left\{(\ce{AgNO3})\right\}}{\pu{230 mL}}= \frac{\pu{0.10 M} \times \pu{105 mL}}{\pu{230 mL}} = \pu{0.0457 M}$$

Similarly, we can also calculate the concentration of $\ce{K2CrO4}$ after dilution using same equation: $$c_f\left\{(\ce{K2CrO4})\right\} = \frac{c_i\left\{(\ce{K2CrO4})\right\} \times v_i\left\{(\ce{K2CrO4})\right\}}{\pu{230 mL}}= \frac{\pu{0.35 M} \times \pu{125 mL}}{\pu{230 mL}} = \pu{0.190 M}$$

Since $\ce{AgNO3 (aq) -> Ag+ (aq) + NO3- (aq)}$, $[\ce{Ag+}] = c_f(\ce{AgNO3}) = \pu{0.0457 M}$.

And also, since $\ce{K2CrO4 (aq) -> 2K+ (aq) + CrO4^2- (aq)}$, $[\ce{CrO4^2-}] = c_f(\ce{K2CrO4}) = \pu{0.190 M}$.

Now, using these final concentrations, you can find $Q_\mathrm{sp}$ (well defined in andselisk's answer elsewhere) that is $\approx 4.0\times 10^{-4}$. The numerical value of $Q_\mathrm{sp}$ is greater than $K_\mathrm{sp}$, and hence $\ce{Ag2CrO4}$ would precipitate.