Convert given concentrations to moles of I鈦 per liter of solution
For each pair, we need to find the concentration of I鈦 ions in moles per liter. Remember to consider the stoichiometry of the compounds.
### Pair (a) ###
(a) \(0.10\,M\, BaI_{2}\) and \(0.25\,M\, KI\).
For \(BaI_2\), every 1 mole will dissociate into 2 moles of \(I^-\) ions:
\[1\, BaI_2 \rightarrow 1\,Ba^{2+} + 2\,I^-\]
For \(0.10\,M\, BaI_{2}\):
\[0.10\,M\,I^-_{(from\,BaI_2)} = 0.10\,M\, BaI_2 \times 2 = 0.20\,M\]
For \(KI\), every 1 mole will dissociate into 1 mole of \(I^-\) ions:
\[1\, KI \rightarrow 1\,K^+ + 1\,I^-\]
For \(0.25\,M\, KI\):
\[0.25\,M\,I^-_{(from\,KI)} = 0.25\,M\, KI \times 1 = 0.25\,M\]
### Pair (b) ###
(b) \(100\,mL\) of \(0.10\,M\, KI\) solution and \(200\,mL\) of \(0.040\, M\, ZnI_{2}\) solution.
For \(KI\):
\[0.10\,M\,I^{-}_{(from\,KI)} = 0.10\,M\, KI \times 1\]
For \(ZnI_2\):
\[1\, ZnI_{2} \rightarrow 1\, Zn^{2+} + 2\, I^-\]
\[0.040\,M\,I^{-}_{(from\,ZnI_2)} = 0.040\,M\, ZnI_{2} \times 2\]
### Pair (c) ###
(c) \(3.2\,M\) HI solution and a solution made by dissolving \(145\,g\) of NaI in water to make \(150\,mL\) of solution,
For \(HI\):
\[3.2\,M\,I^-_{(from\,HI)} = 3.2\,M\, HI \times 1\]
For NaI, we need to calculate the molarity of NaI and then find I鈦 concentration:
\[M_{NaI} = \frac{n_{NaI}}{V_{solution}}\]
- We first calculate the number of moles of NaI (\(n_{NaI}\)) by dividing the mass (145 g) by its molar mass (149.89 g/mol):
\[n_{NaI} = \frac{145\,g}{149.89\,g/mol} = 0.967\,mol\]
- Next, we convert the volume from mL to L:
\[V_{solution} = 150\,mL \times \frac{1\,L}{1000\,mL} = 0.150\,L\]
- Now, we calculate the molarity of NaI:
\[M_{NaI} = \frac{0.967\,mol}{0.150\,L} = 6.45\,M\]
- Finally, we find the concentration of \(I^-\) ions:
\[6.45\,M\,I^-_{(from\,NaI)} = 6.45\,M\, NaI \times 1\]