Question

10.00 mL of 0.10 M NaOH is added to 25.00 mL of 0.10 M HCl during a titration

Answer 1

Answer:

Explanation:

10 mL = .01 L .

25 mL = .025 mL .

10 mL of .1 M NaOH will contain .01 x .1 = .001 moles

25 mL of .1M HCl will contain .025 x .1 = .0025 moles

acid will neutralise and after neutralisation moles of acid remaining

= .0025 - .001 = .0015 moles .

Total volume = .01 + .025 = .035 L

concentration of remaining HCl = .0015 / .035

Option D is correct.

= .042857 M

= 42.857 x 10⁻³ M .

pH = - log [42.857 x 10⁻³]

= 3 - log 42.857

= 3 - 1.632

= 1.368 .