pH of a Buffered Solution Using tabulated Ka and Kb values, calculate the pH of

Question

pH of a Buffered Solution Using tabulated Ka and Kb values, calculate the pH of the following solutions Answer to 2 decimal places. Ka & Kb values may be found in Zumdahl, Chemical Principles 7th ed. Appendix 5 Tables A5.1, A5.3 A solution containing 1.9x10 M HCOOH and 3.2x10 i M HCooNa. 1 pts Submit Answer Tries 0/8 1000.0 mL of solution containing 5.0 g of HNO2 and 20.0 g of NaNO2. 1 pts Submit Answer Tries 0/8 A solution made by mixing 115.00 mL of 1.100 M HOBr with 55.00 mL of 0.950 M NaoH. 1 pts Submit Answer Tries 0/8

Explanation / Answer

we know that

for buffers

pH = pKa + log [salt / acid ]

so

pH = pKa + log [ HCOONa / HCOOH]

pKa of formic acid (HCOOH) is 3.77

so

pH = 3.77 + log [ 3.2 x 10-1 / 1.9 x 10-1]

pH = 3.9964

so

pH of the given solution is 3.9964

2)

in this case

pH = pKa + log [ NaN02 / HN02]

now

moles = mass / molar mass

so

moles of NaN02 = 20 / 69 = 0.2899

moles of HN02 = 5 / 47 = 0.106383

now

pH = pKa + log [NaN02 / HN02]

we know that

concentration = moles / volume

as the volume is same for both

ratio of concentration = ratio of moles

so

pH = 3.398 + log [ 0.2899 / 0.106383]

pH = 3.8333

so

pH of the given solution is 3.8333

3)

now

moles = concentration x volume (ml ) / 1000

so

moles of HOBr = 1.1 x 115 / 1000 = 0.1265

moles of NaOH = 0.95 x 55 / 1000 = 0.05225

now

the reaction is

HOBr + NaOH ---> NaOBr + H20

we can see that

moles of HOBr reacted = moles of NaOH added = 0.05225

moles of NaOBr formed = moles of NaOH added = 0.05225

so finally

moles of HOBr = 0.1265 - 0.05225 = 0.07425

moles of NaOBr = 0.05225

now

pH = pKa + log [NaOBr / HOBr]

pKa of HOBr is 8.65

so

pH = 8.65 + log [ 0.05225 / 0.07425]

pH = 8.4974

so

pH of the given solution is 8.4974

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