Question

The molarity of sulfuric acid in a fully charged car battery is 5.2 m. when fully discharged the molarity is 4.8m. a 2.00ml sample of battery acid was titrated with 31.26ml of
a.621 m sodium hydroxide solution. does the battery need to be recharged

Answer 1

No, the battery does not need to be recharged. The balanced reaction is: NaOH + H2SO4 ==> Na2SO4 + H2O So for each mole of H2SO4, one mole of NaOH is needed. So let's determine the number of moles of NaOH we used: 0.03126 L * 0.621 mol/L = 0.01941246 mol So we now know that 0.01941246 moles of H2SO4 was present in the sample. And since molarity is defined as moles per liter, we can divide the number of moles we had by the number of liters to get molarity. So: 0.01941246 / 0.002 = 9.706 M. This molarity is way too high to be reasonable. So let's do a sanity check on the original measured quantities. Our original sample is 2.00 ml and we titrated with 31.26 ml or 31.26/2 = 15.63 times as much base. So the molarity should be that value times the molarity of the base, which is 15.63 * 0.621 = 9.706 which matches the original calculated figure. The conclusion is that the battery does not need to be recharged. If anything, it's over charged. Also, it's highly likely that this problem has a typo and that the figures given are incorrect. Please check your original problem before using this as your answer.

Answer 2

This question requires utilization of  the equivalence point of a reaction, which is a point in the chemical reaction where the two solutions have been mixed in exactly the right proportions. This means that,

$M_1V_1=M_2V_2$, where $M_1$ is the molarity of the acid sulfuric acid, $V_1$ the volume of acid, $M_2$ is the molarity of the base sodium hydroxide and $V_2$ is the volume of the base.

First we write a balanced equation,

$H_2SO_4 + 2NaOH ==> Na_2SO_4 + 2H_2O$.

The stochiometric ratio of $H_2SO_4: NaOH$ is 1:2. Which means 1 mol $H_2SO_4$ react with 2 mol$NaOH$. Using the volume and molarity of $NaOH$  given we find the molarity of $H_2SO_4$.

$31.26ml \times \frac{1L}{1000ml} \times \frac{0.621mol}{L} = 0.0194mol NaOH$

Using moles of $NaOH$ we find the molarity

$0.0194mol NaOH \times \frac{1mol H_2SO_4}{2mol NaOH} \times \frac{1 mol}{2.0ml} \times \frac{1000ml}{L} = 4.8 M H_2SO_4$

It is fully discharged so the battery needs to be recharged.