At the point of precipitation: $[\text{Ba}^{2+}][\text{SO}_4^{2-}] = K_{sp} = 1 \times 10^{-10}$

Find $[\text{SO}_4^{2-}]$ in the final mixture:

Moles of SO$_4^{2-}$ added $= 0.050\ \text{L} \times 1\ \text{M} = 0.05\ \text{mol}$

Final volume $= 500\ \text{mL} = 0.5\ \text{L}$

$[\text{SO}_4^{2-}]_{\text{final}} = \dfrac{0.05}{0.5} = 0.1\ \text{M}$

Find $[\text{Ba}^{2+}]$ in the final mixture:

$[\text{Ba}^{2+}]_{\text{final}} = \dfrac{K_{sp}}{[\text{SO}_4^{2-}]} = \dfrac{1\times10^{-10}}{0.1} = 1\times10^{-9}\ \text{M}$

Back-calculate original concentration: Final volume is 10× original Ba$^{2+}$ volume (500 mL total, 450 mL original solution).

Original $[\text{Ba}^{2+}] = 1\times10^{-9} \times \dfrac{500}{450} \approx \mathbf{2\times10^{-9}}\ \text{M}$

Let the price of the book $= p$. Total paid $= 1.04p$.

She paid $10 and received change less than $3.00, meaning total paid was more than $7.00:

$1.04p > 7.00 \Rightarrow p > 6.73$

She also received some change, so total paid $< \$10.00$:

$1.04p < 10.00 \Rightarrow p < 9.615$

So $6.73 < p < 9.615$. Now check each statement:

• Statement A: $p < \$9.50$ — Not necessarily, since $p$ can be up to $9.615$. ✗
• Statement B: $p > \$6.90$ — Not necessarily, $p$ can be $6.80$ (giving change $\approx \$2.93 < \$3.00$). ✗
• Statement C: Sales tax $= 0.04p < 0.04 \times 9.615 \approx \$0.385 < \$0.45$. This is always true within the valid range. ✓
• Statement D: Tax $> \$0.50$ would require $p > 12.50$, which is impossible here. ✗

Only Statement C must be true.