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Posted by on Saturday, May 14, 2011 at 10:40am.

A) Calculate the pH of a buffer system that is 0.06 M HNO2 and 0.160 M NaNO2

B) What is the pH after 2.00 M NaOH are added to 1.00 L of this buffer

I can do part A but I don't know how to part B

  • Chem - , Saturday, May 14, 2011 at 1:57pm

    The easiest way is to work up an ICE chart as below. I like to work in millimoles. For the original buffer, if we have 1L of the original buffer, we will have the following millimoles of HNO2 and NaNO2.
    mmoles HNO2 = 0.06M x 1000 mL = 60
    mmoles NaNO2 = 0.160M x 1000 mL = 160

    You don't say how much 2M NaOH is added to the buffer. I will use 1.0 mL of 2M = 2.0 mmoles.

    ..........HNO2 + OH^- ==> NO2^- + H2O
    initial...60......0........160
    added............2.0...........
    change...-2.0....-2.0......+2.0
    equil......58.....0.........162

    Then pH = pKa + log(base)/(acid)
    pH = pKa + log (162/58) = ??

  • Chem - , Saturday, May 14, 2011 at 2:06pm

    By the way, the correct way is to work in M (and not millimoles) so that for (base) we substitute moles/L which will be 162 mmoles/1001 mL = ?? and for (acid) we substitute 58 mmoles/1001 mL. So the HH equation looks like this.
    pH = pKa + log (162/1001)/(58/1001) BUT you notice that the 1001 denominator in each cancels so we are left with 162/58. After working a few thousand of these problems I think in terms of millimoles because I know the volume part will ALWAYS cancel. I mention this because some profs count off for using millimoles and not the "real" concn term. (I did in my classes.) I tried to get the students into the habit of writing
    pH = pKa + log(162/V)/(58/V) and the Vs cancel no matter what V is. That way we don't need to do an extra division for the numerator and another division for the denominator before finding the fraction for mmols base/mmols acid.

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