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Assuming complete dissociation, calculate the pH of the following solutions : (a)0.003 M HCl , (b)0.005 M NaOH , (c)0.002 M HBr , (d)0.002 M KOH |
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Answer» SOLUTION :(a)pH of 0.003 M HCl : Complete ionization of HCl, then solution , [HCl]=`[H^+]`= 0.003 M = `3.0xx10^(-3)` M pH =-log `[H^+]` `=-log (3.0xx10^(-3))={log 3.0+log 10^(-3)}` =-(0.4771-3.0)=-(-2.5229) =+2.5229 `approx` 2.52 (B)pH of 0.005 M NaOH : NaOH is a strong BASE , its complete ionization occurs `[OH^-]`=[NaOH ] =0.005 M = `5.0xx10^(-3)` M So, pOH = -log `[OH^-]` `=-log (5.0xx10^(-3))=-(log 5+ log 10^(-3))` =-0.6990-3.0)=-(-2.3010) =+2.3010 Now, pH = 14.0-pOH =14.0- 2.3010 =11.6990 `approx` 11.70 (c)pH of 0.002 M HBR : HBr is strong acid and 100% ionization occur, `[HBr]=[H^+]=0.002 M = 2.0xx10^(-3)` M `therefore` pH=-log `[H^+]` `=-log (2.0xx10^(-3))=-{log 2.0 + log 10^(-3)}` =-(0.3010-3.00)=-(-2.6990) =+2.699 `approx` 2.70 (d) pH of 0.002 M KOH : KOH is strong base and complete ionization in solution, `[OH^-]=[KOH]=0.002 M =2.0xx10^(-3)` M pOH = -log `[OH^-]` `=-log (2.0xx10^(-3))` `=[log 2.0 + log 10^(-3)]` =-{0.3010 -3.0} = -(-2.6990) =+2.6990 pH = 14.0-pOH = 14.0-2.699 =11.3010 `approx` 11.30 |
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