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If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

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प्रश्न

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

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उत्तर

`["KOH"_("aq")] = 0.561/(1/5) "g"/"L"`

= 2.805 g/L

`= 2.805 xx 1/56.11 "M"`

= 0.05 M

`"KOH"_("aq") -> "K"_("aq")^+ + "OH"_(("aq"))^-`

`["OH"^-] = 0.05 "M" = ["K"^+]`

`["H"^+] ["H"^-] = ["K"^+]`

`["H"^+] "K"_"w"/(["OH"^-])`

`= 10^(-14)/0.05 = 2xx 10^(-13) "M"`

`therefore "pH" = 12.70`

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अध्याय 6: Equilibrium - EXERCISES [पृष्ठ २३७]

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एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
अध्याय 6 Equilibrium
EXERCISES | Q 7.57 | पृष्ठ २३७
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