Question

a solution is made by dissolving 30 g of calcium acetate, ca(ch3coo)2, in enough water to make exactly 250. ml of solution. calculate the molarity of each species: ca(ch3coo)2, ca2+, ch3coo-

Explanation:

Step1: Calculate moles of calcium acetate

The molar mass of $Ca(CH_3COO)_2$ is $M=(40.08)+2\times(12.01 + 3\times1.01+16.00\times2)=158.17\ g/mol$. The number of moles of $Ca(CH_3COO)_2$, $n=\frac{m}{M}$, where $m = 30\ g$. So $n=\frac{30\ g}{158.17\ g/mol}\approx0.19\ mol$.

Step2: Calculate molarity of calcium acetate

The volume of the solution $V = 250\ mL=0.25\ L$. The molarity of $Ca(CH_3COO)_2$, $M_{Ca(CH_3COO)_2}=\frac{n}{V}=\frac{0.19\ mol}{0.25\ L}=0.76\ mol/L$.

Step3: Determine ion - molarities from dissociation

The dissociation of $Ca(CH_3COO)_2$ in water is $Ca(CH_3COO)_2
ightarrow Ca^{2 +}+2CH_3COO^-$. For every mole of $Ca(CH_3COO)_2$ that dissociates, we get 1 mole of $Ca^{2+}$ and 2 moles of $CH_3COO^-$.
The molarity of $Ca^{2+}$ is the same as the molarity of $Ca(CH_3COO)_2$, so $[Ca^{2+}]=0.76\ mol/L$.
The molarity of $CH_3COO^-$ is twice the molarity of $Ca(CH_3COO)_2$, so $[CH_3COO^-]=2\times0.76\ mol/L = 1.52\ mol/L$.

Answer:

$[CH_3COO^-]=1.52\ mol/L$, $[Ca^{2+}]=0.76\ mol/L$, $[Ca(CH_3COO)_2]=0.76\ mol/L$