# Molarity and Ion Concentration: Two Questions?

Jun 27, 2017

$\text{Phosphoric acid}$ is relatively weak....

#### Explanation:

Significant quantities of ${H}_{2} P {O}_{4}^{-}$ and a little $H P {O}_{4}^{2 -}$ would be present at $\text{equilibrium}$......it is NOT a triacid.....For $\text{option B}$, there would be SOME $H S {O}_{4}^{-}$ present at equilibrium, even tho' $\text{bisulfate}$ is moderately strong. Whereas $\text{acetic acid}$ is weak..........

On the other hand, ${K}_{2} S {O}_{4}$ would give 3 equiv of ions......

${K}_{2} S {O}_{4} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} 2 {K}^{+} + S {O}_{4}^{2 -}$

$A$ is thus the best answer.

As for the second problem, your approach is correct......

$2 A g N {O}_{3} \left(a q\right) + C a {I}_{2} \left(a q\right) \rightarrow 2 A g I \left(s\right) \downarrow + C a {\left(N {O}_{3}\right)}_{2} \left(a q\right)$

Silver iodide is as soluble as a brick, and precipitates from solution as a bright yellow powder. Calcium ion remains in solution as a spectator.........

"Concentration of calcium ion"=("Moles of Ca"^(2+))/("Volume of solution")

$= \frac{1.25 \cdot L \times 1.00 \cdot m o l \cdot {L}^{-} 1}{2.00 \cdot L + 1.25 \cdot L} \cong 0.4 \cdot m o l \cdot {L}^{-} 1$