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

Answer:

0.1357 M

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:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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 $H_2SO_4$ :

$Moles=0.1450 \times {10\times 10^{-3}}\ moles$

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