Question

What is the pH of a saturated solution of silver hydroxide, given Ksp = 1.55×10^-8?

Answer 1

`pH\text{ = 10.1}`

Step-by-step explanation:

We start off by writing the ionization equation

We have this as follows:

`AgOH\text{ }\rightarrow\text{ Ag}^+\text{ + OH}^-`

Let us have the concentration of the silver and hydroxide ions as x M

Thus, we have this as:

`\begin{gathered} K_(sp)\text{ = \lbrack Ag}^+]\text{ + \lbrack OH}^-] \\ 1.55\text{ }*\text{ 10}^(-8)\text{ = x}* x \\ x^2\text{ = 1.55 }*\text{ 10}^(-8) \\ x\text{ = }\sqrt{(1.55\text{ }*10^(-8))} \\ x\text{ = 0.0001245} \end{gathered}`

Thus, we have the value of x as 0.0001245

Mathematically:

`\begin{gathered} pOH\text{ = -log\lbrack x\rbrack} \\ pOH\text{ = -log \lparen0.0001245\rparen} \\ pOH\text{ = 3.90} \end{gathered}`

However:

`\begin{gathered} pH\text{ = 14 - pOH} \\ pH\text{ = 14-3.90} \\ pH\text{ = 10.1} \end{gathered}`