On May 26, 1993 reported research showing that at least some of alaskas finer pl

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? 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?

Explanation / Answer

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!

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