Question

12. A chemist needs to bring to an endpoint a 15.4 mL solution of 2.4 M HCI. He has a 1.25 M solution of NaOH

available. How many grams of the NaOH would he need to titrate against the HCI?

Answer 1

8.02 g of NaOH solution

Step-by-step explanation:

The endpoint of a titration is the exact point in which the solution changes colour.

First, we can start off by writing down a balance chemical equation of the situation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(aq)

As shown in the question, the chemist needs a titre of 15.4 mL of 2.4 mol/L HCl to titrate the NaOH. Therefore, titre = 0.0154 L.

We can now find the number of moles of HCl present:

number of moles (n) = concentration (c) ÷ volume (V)

∴ n(HCl) = c/V = 2.4/0.0154 (using the titre) = 155.84 mol

Since stoichiometric ratio of the above equation is 1 : 1, therefore:

n(HCl) = n(NaOH).

∴ n(NaOH) = 155.84 mol

Now we have the number of moles of NaOH, we can rearrange the formula we used before, n = c/v to find the volume required.

v = c/n = 1.25/155.84 = 0.00802 L = 8.02 mL.

Since 1 mL is equal to 1 g, thus, we require 8.02 g of NaOH solution to titrate the HCl.