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If 0.561 g KOH is dissolved in water to give 200 mL of solution at 298 K, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its pH ? |
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Answer» Solution :`[KOH]=(0.561)/(56)xx(1000)/(200)M=0.050M` As `KOH rarr K^(+)+OH^(-), :. [K^(+)]=[OH^(-)]=0.05M` `[H^(+)]=K_(w)//[OH^(-)] = 10^(-14)//0.05 = 10^(-14)//(5xx10^(-2))=2.0xx10^(-13)M`. `pH = - log [H^(+)]=- log (2.0xx10^(-13))=13-0.3010=12.699` |
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