I am having a little touble wiht my pre-lab exercises. I think I am making them

Question

I am having a little touble wiht my pre-lab exercises. I think I am making them harder than they actually are

I can do a. and b. really need help on c through e

7.0 mL of 6.0 m NaOH are diluted with water to a volume of 400 mL .

a.       Calculate the molarity of the NaOH stock solution.

In an acid-base standardization of the above prepared solution, 23.46 mL of the NaOH stock solution are needed to neutralize a 25.00 mL sample of 0.116 HCl solution.

b.      Write the chemical equation for this reaction.

c.       c. Using the titration values, what is the calculated molarity of the NaOH stock solution?

d.      Why would (or should) you expect the answer for part a and c to be different?

e.      What is the molarity of the OH- in the standardized NaOH solution?

Explanation / Answer

a>new concentration = 6*7/400 = 0.105M

b> NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(l)

c> molarity = 0.116*25/23.46 = 0.124M

d> it is because of the fact that heat of mixing and degree of dissociation are not taken into account

e> 0.124M

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.