Respuesta :
Answer:
[tex]pH\text{ = 10.1}[/tex]Explanation:
We start off by writing the ionization equation
We have this as follows:
[tex]AgOH\text{ }\rightarrow\text{ Ag}^+\text{ + OH}^-[/tex]Let us have the concentration of the silver and hydroxide ions as x M
Thus, we have this as:
[tex]\begin{gathered} K_{sp}\text{ = \lbrack Ag}^+]\text{ + \lbrack OH}^-] \\ 1.55\text{ }\times\text{ 10}^{-8}\text{ = x}\times x \\ x^2\text{ = 1.55 }\times\text{ 10}^{-8} \\ x\text{ = }\sqrt{(1.55\text{ }\times10^{-8})} \\ x\text{ = 0.0001245} \end{gathered}[/tex]Thus, we have the value of x as 0.0001245
Mathematically:
[tex]\begin{gathered} pOH\text{ = -log\lbrack x\rbrack} \\ pOH\text{ = -log \lparen0.0001245\rparen} \\ pOH\text{ = 3.90} \end{gathered}[/tex]However:
[tex]\begin{gathered} pH\text{ = 14 - pOH} \\ pH\text{ = 14-3.90} \\ pH\text{ = 10.1} \end{gathered}[/tex]