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PH-calculation question.

  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 23, 2004 #2
    HCl+NaOH------> NaCl(aq) + H2O

    you need 1 mole of NaOH to neutralize 1 mole of HCl according to the balanced equation.

    So you have .102 M HCl which means you have .102 moles HCl/ 1 L of solution so if you have 100 mL of .102 M HCL you have .102 moles HCl/1 L soln x 1L soln/1000mL x 100mL= .0102 moles of HCl. Doing the same thing for NaOH, you have .0078 moles of NaOH. Since NaOH and HCl react in a 1:1 ratio you can obviously see that NaOH is the limiting reagent. Thus .0102moles HCl-.0078 moles HCl =.0024 moles of HCl left. Now I am assuming that you are supposed to assume that 100 mL of HCl+100mL of NaOH=200 mL solution to make things easier, but in reality volumes don't add. Anyway assuming that they do, you have .0024 moles HCl/200mL solution x 1000mL/1L =.012 M HCl so pH=-log[H+]=-log[.012]=1.92
  4. Nov 23, 2004 #3
    Also, HCl is a strong acid so it will dissociate completely
    Last edited: Nov 24, 2004
  5. Nov 24, 2004 #4
    HCl + NaOH --> Na+(aq) + Cl-(aq) + H2O

    Na and Cl wont react and form NaCl because the corresponding base of a strong acid is weak and the corresponding acid of a base is also weak, so the formula for those who react is:

    H+ + OH- --> H20
  6. Nov 25, 2004 #5
    Thats why I wrote NaCl (aq). That implies that the Na and Cl split apart into ions.
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