Ncert Exercises & Solutions

7.48 Assuming complete dissociation, calculate the pH of the following solutions:

(i) 0.003 M HCl (ii) 0.005 M NaOH (iii) 0.002 M HBr (iv) 0.002 M KOH

$(i) 0.003 MHCI :$
$H_{2} O + HCI \leftrightarrow H_{3} O^{+} + {CI}^{-}$
$Since HCI is completely ionized,$
$[H_{3} O^{+}] = [HCI]$
$\Rightarrow [H_{3} O^{+}] = 0.003$
$Now,$
$pH = - \log [H_{3} O^{+}] = - \log (. 003)$
$= 2.52$
$Hence, the pH of the solution is 2.52 .$

$(ii) 0.005 M NaOH : {NaOH}_{(aq)} \leftrightarrow {Na}_{(aq)}^{+} + {HO}_{(aq)}^{-} [{HO}^{-}] = [NaOH] \Rightarrow [{HO}^{-}] = 0.005 pOH = - \log [{HO}^{-}] = - \log (0.005) pOH = 2.30 ∴ pH = 14 - 2.30 = 11.70 Hence, the pH of the solution is 11.70 .$

(iii) 0.002 HBr:

$HBr + H_{2} 0 \leftrightarrow H_{3} O^{+} + {Br}^{-}$
$[H_{3} O^{+}] = [HBr]$
$\Rightarrow [H_{3} O^{+}] = . 002$
$∴ pH = - \log [H_{3} O^{+}]$
$= - \log (0.002)$
$= 2.69$

Hence, the pH of the solution is 2.69

(iv) 0.002 M KOH:

${KOH}_{(aq)} \leftrightarrow K_{(aq)}^{+} + {OH}_{(aq)}^{-}$
$[{OH}^{-}] = [KOH]$
$\Rightarrow [{OH}^{-}] = . 002$
$Now, pOH = - \log [{OH}^{-}]$
$= 2.69$
$∴ pH = 14 - 2.69$
$= 11.31$

Hence, the pH of the solution is 11.31

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