1. **Problem statement:** We want to find the concentration of CO_2 in ppm that causes the ocean pH to drop from 8.3 to 7.8. 2. **Given relation:** $$[H^+] = \sqrt[3]{\frac{K_1^2 K_2 P_{CO_2}^2}{2 K_H^2 K_{sp}}}$$ 3. **Recall pH definition:** $$pH = -\log_{10} [H^+]$$ 4. **Calculate initial [H^+] at pH 8.3:** $$[H^+]_{8.3} = 10^{-8.3}$$ 5. **Calculate new [H^+] at pH 7.8:** $$[H^+]_{7.8} = 10^{-7.8}$$ 6. **From the relation, since all constants are fixed,** $$[H^+] \propto P_{CO_2}^{2/3}$$ 7. **Set up ratio:** $$\frac{[H^+]_{7.8}}{[H^+]_{8.3}} = \left(\frac{P_{CO_2,new}}{P_{CO_2,old}}\right)^{2/3}$$ 8. **Plug in values:** $$\frac{10^{-7.8}}{10^{-8.3}} = \left(\frac{P_{CO_2,new}}{420}\right)^{2/3}$$ 9. **Simplify left side:** $$10^{8.3 - 7.8} = 10^{0.5} = \sqrt{10} \approx 3.162$$ 10. **Raise both sides to the power of $\frac{3}{2}$ to solve for $\frac{P_{CO_2,new}}{420}$:** $$\left(3.162\right)^{\frac{3}{2}} = \frac{P_{CO_2,new}}{420}$$ 11. **Calculate:** $$3.162^{1.5} = 3.162 \times \sqrt{3.162} \approx 3.162 \times 1.778 = 5.623$$ 12. **Find new CO_2 concentration:** $$P_{CO_2,new} = 420 \times 5.623 = 2360.7$$ 13. **Report answer to 2 significant figures:** $$\boxed{2400}$$ ppm This matches the given approximate answer and shows the CO_2 concentration needed to cause a half unit drop in pH.