1.Determine the pH (to two decimal places) of a solution prepared by adding 51.3
ID: 1003246 • Letter: 1
Question
1.Determine the pH (to two decimal places) of a solution prepared by adding 51.3 mL of 0.0915 M sodium hydrogen sulfite (NaHSO3) to 39.4 g of sodium sulfite (Na2SO3). Assume the volume of the solution does not change and that the 5% approximation is valid.
2.Calculate the pH (to two decimal places) of the buffer solution after the addition of 1.15 g of sodium butanoate (NaC3H7COO) to the buffer solution above.
Assume 5% approximation is valid and that the volume of solution does not change.
3. The mole fraction of an aqueous solution of ammonium nitrite is 0.494. Calculate the molarity (in mol/L) of the ammonium nitrite solution, if the density of the solution is 1.45 g mL-1.
4. Determine the mole fraction of calcium nitrite in a 9.28 M aqueous solution of calcium nitrite. The density of the solution is 1.47 g mL-1.
Please answer the following 4 questions. Will upvote
Explanation / Answer
Solution:
1) NaHSO3 --> 0.0513 * 0.0915 = 0.00469 moles of NaHSO3
Na2SO3--> Concentration = (39.4/126)/0.0513 = 6.0954 M
by using Hendersen-haselbalch equation,
pH = pK +log {[conj.base]/[conj.acid]}
= 7.2 + log {6.0954/0.0915}
= 9.0235
2) [sodium butanoate] = (1.15/110)/1 = 0.01045
pH = -log (0.01045) = 1.980
3) first assume 100 g of solution
49.4 gm of ammounium nitrate
50.6 gm of water
moles of solute --> 49.4 g * (1 mol / 80 g ) = 0.6175 moles
kg of solvent ------> 50.6 g * (1 kg / 1000g) = 0.0506 kg
molarity = 0.6175 moles / 0.0506 kg = 12.2 M
4) Assume the amount of solution with 1 kg of water ---> 9.28 M of calcium nitrate (aq)
moles of water ----> 1 kg * (1000g / 1kg) * ( 1mol/18.06 g) = 55.37 moles of water
mole fraction of calcium nitrate = 9.28 / (9.28 + 55.37) = 0.1435
mole fraction of water = 1 - 0.1435 = 0.8565
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