[Lindsay]

Hello,
I need some help with various problems. Please make the answers are correct

1. A 4 g mass of NaOH is dissolved in 5mL of water
a. what is the approximate molar concentration of the NaOH?
My answer..
4 g X 1 mol NaOH/40.00 g NaOH = .1 mol NaOH

M=m/V .1 mol NaOH/.005L = 20 M

b. a 4 mL aliquot of this solution is diluted to 500 mL of solution. What is the approximate molar concentration of NaOH in the diluted solution?
m=M X V
20M X .004 = .08 m NaOH
M= .08m/.5L = .16 M

c. Calculate the mass of KHC8H4O4 that reacts with 15 mL of the NaOH solution

KHC8H4O4 + NaOH ----- H2O + NaKC8H4O4

15 mL/1000 X .16 M NaOH/ 1L X 1 mol KHC8H4O4/1 mol NaOH X 204.44 g KHC8H4O4/ 1 mol KHC8H4O4 = 0.49 g KHC8H4O4

Thanks,

Lindsay

[Astrokel]

Correct! well done just a side note:

m=M X V
20M X .004 = .08 m NaOH

i don't know if they will penalized you for using m instead of n (m = M X V) as we usually use m for mass and n stands for moles. But your SI unit for moles is wrong it is not m, it is in mol

[Lindsay]

Thanks!
I have two more problems.
Here they are

1. (A) 0.4040 g sample of potassium hydrogen phthalate, KHC8H4O4 ( molar mass = 204.44) is dissolved with  50 mL of deionized water in a 125 mL flask. The sample is titrated to the phenolphthalein endpoint with 14.71 mL of a sodium hydroxide solution. What is the molar concentration of the NaOH solution?
My answer is

0.4040 g KHC8H4O4 X 1 mol KHC8H4O4/204.44 g KHC8H4O4 X 1 mol NaOH/ 1 mol KHC8H4O4 = 0.001976 mole of NaOH

M= Moles of Solute/L of solution

M = 0.001976/ 0.01471 L = 0.1343 M of NaOH

(B) A 25.000 mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125 mL flask and 2 drops of phenolphthalein are added. The above NaOH solution (the titrant) is used to titrate the nitric acid solution (the analyte). If 18.92 mL of the titrant is dispensed from a buret in causing a color change of the phenolphthalein, what is the molar concentration of the nitric acid solution?

I'm not sure on this one. The 2 drops are throwing me off.

you have a reaction of
HNO3 + NaOH----H2O + NaNO3
 so
18.92/1000 X 0.1343/1 L Solution X 1 mol HNO3/ 1 mol NaOH = 0.002541 mole of HNO3

Then, I guess, you need to find the molar conc of the acid

M = moles of solute/ Liters of Solution

M = 0.002541/ 0.25 L = 0.01016

I'm not sure about this two problems. Please advice.

Thanks,

Lindsay

[Astrokel]

Both are correct! 

The 2 drops added are just indicator and it doesn't affect your end point.