Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha
1 answer:
You might be interested in
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