Question

On May 26, 1993 reported research showing that at least some of alaskas finer placer good was--bug sweat. This was published in a popular science news outlet in Alaska. What they found was for one particular strain of bacteria in bugs had been processing gold from ore most efficiently a pH of 3.30 but where ineffective at pHs of above 4 or below 2.7 in order to protect these tiny workers the ore processing vats have a concentration of .0685 M HNO2(aq) and .0815 M NaNO2(aq) what is the pH of the solution? As well if someone where to add NaOh of 26.0 L which is 4M to each vat keeping in mind that vat only holds 2.50x10^3 l of solution what would be the final pH?

Answer 1

1)

pH of solution if

M1 = 0.0685

M2 = 0.0815 NanO2

This is a buffer, use the Henderson Hasselbach equation to model this

pH = pka + log([salt]/[acid])

pKa of Nitrous Acid from databases = 3.39

pH = 3.39 +log([salt]/[acid])

pH = 3.39 + log(0.0815/0.0685) = 3.39 + 0.075 = 3.46

Ph = 3.46

2) Ph if adgin NaOH

V = 26 L

M = 4 M

Let us calculate the amount of moles of NaOH there

M1*V1 = 26*4 = 104 moles of NaOH

Assume that:

Vats were full therefore V = 2500 L

calculate amount of HNO2 and NaNO2

M2V2 = 0.0685*2500 = 171.25 mol of HNO2

M3*V3 = 0.0815*2500 = 203.75 mol of NaNO2

NOTE: I will asume that those 26 liters are not so relevant to the total of 2500 L

Therefore...

171.25 mol of HNO2 react 1:1 with 104 moles of NaOH

67.5 mol of HNO2 are left...

It is still a buffer so let us recalculate concnetration of HNO2

M = moles / V = 67.5 mol of HNO2 / 2500 L = 0.027 M

The molarity of the salt stays the same, it does not react

Apply once again Henderson Haselbach equaiton

pH = pKa +log([salt]/[acid])

pH = 3.39 +log([salt]/[acid])

pH = 3.39 +log([0.0815]/[0.027]) = 3.39 + 0.4797 = 3.869

pH = 3.87 so it will still work... The idea of adding a buffer is pretty awesome since a lot of NaOH did not chang the pH that much!