Answer
a. $10^{-2.3}\approx5.0\times10^{-3}$ mol/L
b. $10^{-1}=0.1$ mol/L.
c. $10^{1.3}\approx20$ times greater.
Work Step by Step
a. Given the equation $pH = -log (x)$, with $pH=2.3$, we have $-log(x)=2.3$; thus $x_1=10^{-2.3}\approx5.0\times10^{-3}$ mol/L
b. With $pH=1$, we have $-log(x)=1$, $x_2=10^{-1}=0.1$ mol/L. With $pH=3$, we have $-log(x)=3$, $x_2=10^{-3}=0.001$ mol/L. Thus the most acidic stomach is $pH=1, x_2=10^{-1}=0.1$
c. Taking the ratio of the above results, we have $\frac{x_2}{x_1}=10^{2.3-1}=10^{1.3}\approx20$ times greater.
