need help with the pre lab. Thanks Sooo much is determined. Strong acid and stro
ID: 1038242 • Letter: N
Question
need help with the pre lab. Thanks Sooo much
is determined. Strong acid and strong base are added to each buffer in a series of steps and the pH is determined after each addition. The resulting pior ea b pH values after each addition will be compared to the calculated pH values for each buffer. Pre-Lab Questions 1. How many grams of sodium acetate (molar mass 82.03 g/mol) must be added to 1.00 L of a 0.200 M acetic acid solution to form a buffer of 4.20? K, value for acetic acid is 1.8-x-10 2. Three milliliters of a 2.0 M solution of HCl are added to 1 liter of buffer solution containing 0.40 moles of the weak acid, propanoic acid (K, 14 x 10) and 0.50 moles of its conju- gate base, sodium propanate. a. What is the original pH of the buffer before the strong acid is added? b. What is the pH of the buffer after the HCl is added? Assume negligible volume change. 3. The weak base-conjugate acid buffer used in this laboratory consists of a weak base ammonia, NHs, and its conjugate acid ammonium chloride, NH.Cl. If the NH, concentration is 0.05 M and the NHCI concentration is 0.05 M, what is the pH of the buffer? K, for NH, is 1.8-x-105.Explanation / Answer
1)
given pH of buffer = 4.2
ka of acid = 1.8x10-5 that is pKa = 4.75
[acid] = 0.2
[acetate] = ?
Now the pH of a buffer is given by Hendersen equation as
pH = pKa + log [conjugate base]/[acid]
4.20 = 4.75 + log [acetate]/0.02
Solving
[acetate]=5.636x10-3 M
thus mass of sodium acetate to be added to 1L of buffer = moles x molar mass
=5.636x10-3 M x 82.03
=0.4623 g /L
2)
a)
pKa of acid = -log ( 1.4x10-5) = 4.85
the pH of buffer originally
pH = pKa + log [conjugate base]/[acid]
= 4.85 + log 0.50/0.40
= 4.95
b)after addition of 3mL 2.00 M HCl
A- + HCl ------------------> HA + Cl-
1000x 0.5 0 1000x 0.4 - initial mmoles
- 3 x2 - - change
494 0 406 - equilibrium mmoles
Thus the pH of this new buffer is
pH = 4.85 + log 494/406
=4.935
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