Answer:
Step-by-step 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 .