Question

use the ph interactive to complete the table.

answer bank

Explanation:

Step1: Calculate \([H^{+}]\) for \(pH = 9\)

Using the formula \(pH=-\log[H^{+}]\), if \(pH = 9\), then \([H^{+}]=10^{-pH}=10^{-9}\)

Step2: Calculate \([OH^{-}]\) for \(pH = 9\)

We know that \(K_w=[H^{+}][OH^{-}]=10^{-14}\). If \([H^{+}]=10^{-9}\), then \([OH^{-}]=\frac{K_w}{[H^{+}]}=\frac{10^{-14}}{10^{-9}} = 10^{-5}\)

Step3: Calculate \(pH\) for \([H^{+}]=10^{-7}\)

Using \(pH =-\log[H^{+}]\), if \([H^{+}]=10^{-7}\), then \(pH=-\log(10^{-7})=7\)

Step4: Calculate \([OH^{-}]\) for \([H^{+}]=10^{-7}\)

Since \(K_w = 10^{-14}\), \([OH^{-}]=\frac{K_w}{[H^{+}]}=\frac{10^{-14}}{10^{-7}}=10^{-7}\)

Step5: Calculate \(pH\) for \([OH^{-}]=10^{-9}\)

First, find \([H^{+}]=\frac{K_w}{[OH^{-}]}=\frac{10^{-14}}{10^{-9}}=10^{-5}\). Then, using \(pH=-\log[H^{+}]\), \(pH =-\log(10^{-5})=5\)

Answer:

pH	\([H^{+}]\) (M)	\([OH^{-}]\) (M)
7	\(10^{-7}\)	\(10^{-7}\)
5	\(10^{-5}\)	\(10^{-9}\)