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

Which expression shows how to determine the Ht

concentration of the solution now?

A. 0.0025/0.025

B. 0.0025/0.035

C. O. 0015/0.025

D. 0.0015/0.035

What is the pH of the solution after this addition?

(Write answer to two decimal places.)

pH =

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.