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 - DrBob222, 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
Then pH = pKa + log(base)/(acid)
pH = pKa + log (162/58) = ??
Chem - DrBob222, 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.