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9 July, 14:35

Calculate the ph of the resulting solution if 25.0 ml of 0.250 m hcl (aq) is added to (b) 35.0 ml of 0.300 m naoh (aq).

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  1. 9 July, 14:46
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    In this question, the solution have 25.0 ml of 0.250 m hcl (aq). The amount of HCl in mole would be: 0.25 mol/1000ml * 25ml = 0.00625 mol

    The solution then added by 35.0 ml of 0.300 m naoh (aq). The NaOH content would be: 0.3mol/1000ml * 35ml = 0.0105 mol

    If you add them up, the number of NaOH remain after reaction would be: 0.0105 - 0.00625 = 0.00425mol

    The total solution volume would be: 25ml+35ml = 60ml = 0.06liter

    The concentration of the OH - ion would be: 0.00425mol / 0.06l = 0.07083 molar

    The pH would be: 14 - log 0.07083 molar = 12.86
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