Problem

Problem

What is Kb for the aqueous fluoride ion given that Kw is 1.0×10−141.0 \times 10^{-14} and Ka for HF is 6.8×10−46.8 \times 10^{-4} at 298 K? a. 16.8×10−4\frac{1}{6.8 \times 10^{-4}} b. (6.8×10−4)(1.0×10−14)(6.8 \times 10^{-4})(1.0 \times 10^{-14}) c. 1.0×10−146.8×10−4\frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} d. 6.8×10−46.8 \times 10^{-4}

Browse the problem library