How does it work when we add acid, HCI? What happens when we add base. NaOH? The
ID: 961293 • Letter: H
Question
How does it work when we add acid, HCI? What happens when we add base. NaOH? The OH^- has added Generalized by the AF out pressed in the OH^- + HF rightarrow F^- + H_2 O Use the Henderson Hasselbalch equation to calculate the pH of a buffer solution prepared from equal volumes of equimolar solutions of H_2 S and NaHS. K_e1 for H_2 S is.1 times 10^-2. H_2 S + NcOH rightarrow NeHS + H_O H_2 H^+ 5 H Ke1 = 1.1 times 0^-7 p^4 = p^ke + 1og [base]/[add] p^4 = pk_e1 + log [N_2 HC]/[H_2 S] pk_1 = -log.1 times 10^-2 = 6.96 Given [NaHS] = [H_2 S] log [NaHS]/[H_2 S] = log [NaHS]/[H_2 S] = log 1= Use the Henderson-Hasselbalch equation to perform the following calculations that you will need for the experiment. The K_2 of acetic acid is 1.8 times 10^-1. Calculate the mass of solid sodium acetate required to mix with 100.0 ml. of 0.10 M acetic acid to prepare a pH 4 buffer. Calculate the mass of solid sodium acetate required to mix with 100.0 ml. of 0.10 M acetic acid to prepare a pH 4 buffer.Explanation / Answer
pH= pKa + log[acetate]/[acid]
The pKa of acetic acid is 4.76, and the pH of buffer we need is 4,
4= 4.76 + log [acetate]/[acid]
-0.76 = log [acetate]/[acid]
if we do the antilog of both sides, we get:
0.174 = [acetate]/[acid]
We know the concentration of acetic acid in the solution, and since you are not changing the volume at all, that's going to stay the same, so :
0.174 = [acetate]/0.1
[acetate]= 0.174*0.1 = 0.0174 M
Now the concentration of acetate in moles/L, we know that the volume if 100 mL or 0.100 L, so we can set up this equation:
0.0174 M= moles/0.100L
moles=0 .00174 moles sodium acetate needed
Mass of acetate = 0.00174 moles * 82.03 g/mol = 0.143 g sodium acetate
I can't find the difference between Q.A & B
Both are sam i think so
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