A lead storage cell contains $500 mL$ of $5.00 M$ $H X_{2} SO X_{4}$ . The cell delivers a steady current of $3.00 A$ for $9650 s$ . Neglecting any change in volume, find the molarity of $H X_{2} SO X_{4}$ after discharge (two decimals). Take $F = 96500 C mo l^{- 1}$ .

Numerical answer type, answer in M, tolerance 0.02.

Answer: 4.4 M

Charge

Q = 3.00 \times 9650 = 28950 C = 0.300 mol

of electrons. Discharge reaction

Pb + PbO X_{2} + 2 H X_{2} SO X_{4} 2 PbSO X_{4} + 2 H X_{2} O

involves

2

electrons per

2 mol

H X_{2} SO X_{4}

, i.e.

1 mol

of acid per mole of electrons, so

0.300 mol

H X_{2} SO X_{4}

is consumed. Initial acid

= 0.500 \times 5.00 = 2.50 mol

; remaining

= 2.20 mol

0.500 L

, so molarity

= 4.40 M