Answer
a. $10^{-5.6}\approx2.5\times10^{-6}$ mol/L
b. $10^{-2.4}\approx4.0\times10^{-3}$ mol/L
c. $10^{3.2}\approx1585$ times greater.
Work Step by Step
a. Given the equation $pH = -log (x)$, with $pH=5.6$, we have $-log(x)=5.6$; thus $x_1=10^{-5.6}\approx2.5\times10^{-6}$ mol/L
b. With $pH=2.4$, we have $-log(x)=2.4$; thus $x_2=10^{-2.4}\approx4.0\times10^{-3}$ mol/L
c. Taking the ratio of the above results, we have $\frac{x_2}{x_1}=10^{5.6-2.4}=10^{3.2}\approx1585$ times greater.