Sodium hydroxide is a strong base. Pure aqueous sodium hydroxide contains only 5

Question

Sodium hydroxide is a strong base. Pure aqueous sodium hydroxide contains only 50 wt% sodium hydroxide. the rest is water. the density of a 50 wt% sodium hydroxide is 1.8g/ml at 25°c.

Formic acid is weak. The "pure" aqueous solution containing 90 wt% formic acid. The density of Formic acid is 1.02 g/ml at 25°c. The ka of the formic acid is 1.8x10-4.

suppose you titration 5.00 vol% aqueous sodium hydroxide solution in a 30.00ml of a 7.00 vol% aqueous formic acid solution until equivalence.

1.Draw a reasonable graph of pH ( yaxis) vs NaOH solution added (xaxis). i. e a generic graph for titration of a strong base into a week acid.
2. calculate PH at equivalence. report value to three significant figures.

Explanation / Answer

HCOOH + NaOH HCOONa + H2O
at the equivalence point,

1 mol NaOH reacts with 1 mol HCOOH to produce 1 mol HCOONa

30 ml of HCOOH means 0.03 mol HCOOH

similarly , NaOH is 0.10 mol NaOH

Molarity of each:
HCOONa = 0.1 M
HCOOH = 0.03 M

the equation is

NaOH (aq) + HCOOH (aq) = HCOONa (s) + H2O (l)

density of NaOH is 1.8 gm/ml

50 wt% of NaOH

density of HCOOH is 1.02 gm/ml

90 wt% of HCOOH

Molar mass HCOOH = 46g/mol
5.76g HCOOH = 5.76/46 = 0.125 mol HCOOH


pH cis alculated using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])
pKa = -log (1.80*10^-4) = 3.75
pH = 3.74+log ( 0.1/0.03)
pH = 3.74 + log 3
pH = 3.74 + 0.477
pH = 4.217

pH at equivalenace point is 4.217

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