Question

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

(a) Write a balanced reaction equation for the acid-base neutralization.
(b) Calculate the molarity of NaOH based on the results of this titration.

Answer 1

0.1357 M

Step-by-step explanation:

(a) The balanced reaction is shown below as:

`2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O`

(b) Moles of
`H_2SO_4` can be calculated as:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For
`H_2SO_4` :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of
`H_2SO_4` :

`Moles=0.1450 * {10* 10^(-3)}\ moles`

Moles of
`H_2SO_4` = 0.00145 moles

From the reaction,

1 mole of
`H_2SO_4` react with 2 moles of NaOH

0.00145 mole of
`H_2SO_4` react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M