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 =

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 .

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :