How did the rate of the reaction change over time in the control reaction?
1 answer:
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Before addition of HCl,
conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M
After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔ CH3COOH + NaCl
Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)
Now, conc. of H+ ions = ka![\frac{[acid]}{[conjugated base]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bacid%5D%7D%7B%5Bconjugated%20base%5D%7D)
where ka = dissociation constant for acid = 10^-5 for Ch3COOH
∴ conc. of H+ ions =
= 1.1277 x 10^-5
Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94
conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M
After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔ CH3COOH + NaCl
Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)
Now, conc. of H+ ions = ka
where ka = dissociation constant for acid = 10^-5 for Ch3COOH
∴ conc. of H+ ions =
= 1.1277 x 10^-5
Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94
Answer : It takes time for the concentration to decrease to 0.100 M is, 22.4 s
Explanation :
Formula used to calculate the rate constant for zero order reaction.
The expression used is:
where,
= initial concentration = 0.537 M
= final concentration = 0.100 M
t = time = ?
k = rate constant = 0.075 M/s
Now put all the given values in the above expression, we get:
Therefore, it takes time for the concentration to decrease to 0.100 M is, 22.4 s