Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [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,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.

where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

The oxidation state of Pb in
is 2. So, moles deposited by Pb is as follows.

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.