Question

What is the pH of a 3.9×10-2 M solution of KOH at 25 °C?

Answer 1

★$\color{darkblue}\underline{\underline{\sf Given-}}$

• Concentration of KOH = ${\sf 3.9×10^{-2}}$

★$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• pH of KOH

★$\color{darkblue}\underline{\underline{\sf Formula \: Used-}}$

$\color{violet}\underline{\boxed{\sf pOH=-log[OH^-]}}$

$\color{violet}\underline{\boxed{\sf pH+pOH=14}}$

★$\color{darkblue}\underline{\underline{\sf Solution-}}$

$\implies{\sf (OH^-)=39×10^{-3}}$

$\implies{\sf pOH = -log[OH^-]}$

$\implies{\sf pOH = -log[39×10^{-3}]}$

$\implies{\sf pOH = 3log39}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀log 39 = 1.59

$\implies{\sf pOH = 3×1.59}$

$\implies{\sf pOH= 4.77}$

★$\underline{\boxed{\sf pH+pOH=14}}$

$\implies{\sf pH+4.77=14}$

$\implies{\sf pH=14-4.77 }$

$\implies{\sf pH=9.23}$

★$\color{darkblue}\underline{\underline{\sf Answer-}}$

pH of KOH is $\color{red}{\sf 9.23}$