The hair on Jamal’s arms is standing straight up.
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: Molarity of the prepared solution is
Explanation:
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
where,
n= moles of solute
= volume of solution in ml = 500 ml
moles of solute =
Now put all the given values in the formula of molarity, we get
Thus molarity of the prepared solution is