Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 408: 56

Increase by a factor of $10$

We are given the model: $pH=2$ $[p H]=-\log [H^{+}] \implies 2=-\log [H^{+}]_{initial}$ $[H^{+}]_{initial}= 0.01$ $[H^{+}]_{after}= 0.1$ $\dfrac{[H^{+}]_{initial}}{[H^{+}]_{after}} =\dfrac{0.1}{0.01}$ $=10$ Our answer is: Increase by a factor of $10$.