Henderson–<span>Hasselbalch equation is given as,
pH = pKa + log [A</span>⁻] / [HA] -------- (1)
Solution:
Convert Ka into pKa
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.54 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.54 - 3.863
log [lactate] / [lactic acid] = 0.677
Taking Anti log,
[lactate] / [lactic acid] = 4.75
Result:
4.75 M lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.54.
NH3(aq) + HCl(aq) ⇒ NH4Cl(aq) >>> (1)
∵ C = n/V; C= concentration, n= No. of moles, and V= volume (L)
∴ n = C*V, n(HCl) = 0.050*(50/1000) = 0.0025 moles
n(NH3) = 0.050*(50/1000) = 0.0025 moles
So, the limiting no. of moles is 0.0025 moles >>> (2)
∵ NH3 is weak base, and HCl is strong Acid (and have the same number of moles) >>> So, without any calculation we can notice that the formed salt (NH4Cl) is acidic salt and the pH is less than 7.
From (1) and (2), The no. of moles of NH4Cl is 0.0025 moles >>> (3)
∴ the concentration of [NH4Cl] = 0.0025 / (total volume per L)
= 0.0025 / ((50 + 50) / 1000) = 0.025 M
NH4+(aq) ⇔ NH3(aq) + H+(aq) >>> (4)
(0.025 - x) (x) (x) >>> (5)
∵ Ka = [NH3] [H+] / [NH4+] >>>> (6)
Ka = Kw / Kb, Kb = 1.8 * 10^-5 >>> (7)
∴ Ka = 10^-14 / 1.8*10^-5 = 5.56*10^-10 >>> (8)
From (4), (5), (7) and (8)
Ka = 5.56*10^-10 = (x * x) / (0.025-x) , we will assume that (0.025 - x) = 0.025
∴ x^2 = (5.56*10^-10)(0.025) = 139*10^-13
∴ x = 3.73*10^-6 = [H+]
∵ pH = - log [H+]
∴ pH = - log 3.73*10^-6 = 5.43
Answer:
True
Explanation:
The arrangement of three groups COOH, CO, OH in the order of reducing boiling point is as follow -
COOH > OH > CO
COOH gets strongly polarised due to the presence of electron withdrawing carboxy group and hence have strong H+ bonds as compared to that of alcohol.
Hence the given statement is true.