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Question regarding PH-value!

  1. Nov 23, 2004 #1
    I have calculated the PH-value for this following solution:

    100 mL 0,102 M HCL and 100 mL 0,0780 M NaOH.

    To calculate the pH in the this solution first I must calculate the number of moles [itex]n_{[H_3O^+]}[/itex].

    [itex]n_{[H_3 O^{+}]} = 0,100 L \cdot 0,0102 \ mol/L + 0,100 L \cdot 0,0780 mol/L = 0,018 mol[/itex]

    This means that [itex][H_3 O^{+}] = \frac{0,018 mol}{0,200 L} = 0,09 mol/L[/itex]

    pH for the solution is then [itex]pH = \textrm{-log}(0,09) = 1,05[/itex]

    I would appriciate if somebody would look at my calculation and then tell me if its accurate ??

    Many thanks in advance.


  2. jcsd
  3. Nov 24, 2004 #2
    No. of moles of HCl in 100mL, 0.102 M soln. = 0.0102
    No. of moles of NaOH in 100mL, 0.078 M soln. = 0.0078

    So, a neutralization takes place and the number of moles of HCl that remain = 0.0102-0.0078=0.0024.

    The concentration of H+ ions is 0.0024/0.200 = 0.012M

    So, pH=-log(0.012)=1.92

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