Homework Help: Chemistry

Posted by Cassie on Saturday, April 9, 2011 at 2:54pm.

The pH of a weak monoprotic acid, HA, is 4.55. It took 39.22 ml of 0.2334 M NaOH to titrate 25.00 ml of the acid.

a. Write an equation for the above reaction.

b. Calculate the molarity of the weak acid

c. Write the equilibrium equation. Construct a table showing initial concentration, change in concentration and the equilibrium concentration. Calculate the ionization constant Ka, for the above acid

d. 23.55 ml of the NaOH were added to partially neutralize a new 25.00 ml sample of the acid. The pH of the mixture was measured as 5.33. Calculate Ka for the acid, HA, using the partial neautralization method

Chemistry - DrBob222, Sunday, April 10, 2011 at 1:22am
a) HA + NaOH ==> H2O + NaA
b) moles NaOH = M x L = ??
Using the coefficients in the balanced equation, convert moles NaOH to moles HA.
Then M HA = moles HA/L HA. You have moles and L, solve for HA.

c) Ka = (H^+)(A^-)/(HA)
millimoles HA = mL x M = 25.00*0.3662 = 9.154
mmoles NaOH = 9.154
I don't know what you want for the chart.
d)I don't know what the "partial neutralization" procedure is. I would use the Henderson-Hasselbalch equation.
..............HA + NaOH ==> NaA + H2O
I = initial. 25.00 x 0.3662M = mmoles.
add NaOH = 23.55 x 0.2334 = mmoles.
mmoles NaA = 23.55 x 0.2334
mmoles HA = initial mmoles - mmoles NaOH . Then pH = 5.33 = pKa + log(base/acid) and solve for pKa, then Ka.

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