7.48. 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.
7.48. 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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1 Answer
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(a) 0.003 M HCl
[H3? O+] = [HCl] = 0.003MpH = −log [H+] = −log (3.0×10−3) = 2.523
(b) 0.005 M NaOH
[OH−] = [NaOH] = 0.005M
[H+] = Kw? / [OH−]? = 10−14/ 0.005? =2×10−12pH= −log [H+]=−log (2×10−12)=11.699
(c) 0.002M HBr
[H+]= [HBr]=0.002pH= −log [H+]=−log0.002=2.699
(d) 0.002M KOH
[OH−]= [KOH]=0.002M
[H+]= Kw / [OH−]? =10−14 / 0.002? =5×10−12pH= −log [H+]=−log (5×10−12)=11.301
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