Question

Please Provide detailed explaination with equations and fulldetailed solutions....

Also, pretty positive that my answers to question 1-2 A are correct, so my only concern and what I need help with is the remainder of the questions, although feel free to verify my answers...Thank you!

1) 7.0 mL of 6.0M NaOH are diluted with water to avolume of 400mL. You are asked to find the molarity of theresulting solution.
a. First find out how many moles ofNaOH there are in 7.0 mL of 6.0 M NaOH. Note: thevolume must be in liters.

No . of moles of NaOH , n = Molarity * Volume in L

= 6.0 M * 0.007 L

= 0.042 moles

b. Since the total number of molesof NaOH is not changed on dilution, the molarity afterdilution can also be found, using the final volume of thesolution. Calculate the molarity.
MV = M'V'

6.0 M * 0.007L = M * 0.4 L

M = 0.105M

2) In an acid-base titration, 22.13 mL of an NaOH solutionare needed to neutralize 24.65 mL of a 0.1094 M HCl solution. To find the molarity of the NaOH solution, we can use the followingprocedure:

MV =M'V'

M * 0.02213 L = 0.1094 * 0.02465 L

M = 0.1218 M

b. Find M_OH- in the NaOHsolution.
_______________M

c. Obtain M _NaOH from M_OH- .
_______________M

3) A 0.2678g sample of an unknown acid requires 27.21 mL of0.1164 M NaOH for neutralization to a phenolphalein endpoint. There are 0.35 mL of 0.1012 M HCl used forback-titration.

a. How many moles of OH^- are used? How many moles of H+ from HCl?

___________moles OH^- ____________molesH⁺

b. How many moles of H⁺ are there in thesolid acid?
____________moles H⁺ in solid

c. What is the molar mass of the unknownacid?
_______________ g