Question

from the given data sheet, can anyone please let me know how to solve these post lab questions ?

Post-Lab Questions 1. Calculate the volume of Solution A that you would need to make 50.00 mL of NaOH solution with the same concentration as solution F. Given that one drop of a liquid is 0.05 ml, how many drops of Solution A would be needed? 2. In general, what effect would diluting a NaOH solution have on the pH of the diluted solution? If the dilution process was continued indefinitely, what would the pH of the NaOH solution eventually reach?

Answer 1

ANSWER:

First, it is needed to complete the data sheet:

• Calculate the concentration of NaOH in solution A

$moles\, NaOH=\frac{mass\, NaOH}{MW\, NaOH}=\frac{0.12\, g}{39.997\, g/mol}=\mathbf{3.00x10^{-3}\, mol}$

$[NaOH]=\frac{moles\, NaOH}{V_{sol}}=\frac{3.00x10^{-3}\, mol}{0.25\, ml}=\mathbf{0.012\, M}$

• Fill the table for solution B-F
  • The moles of NaOH are the same before dilution (concentrated solution) and after dilution (diluted solution. It is calculated with the equation:

$moles\, NaOH=[NaOH]_{conc}\, (mol/L)\times V_{conc}\, (L)$

• The final concentration is calculated with the following equation

$[NaOH]_{conc}\times V_{conc} =[NaOH]_{dil}\times V_{dil}$

$[NaOH]_{dil} =\frac{[NaOH]_{conc}\times V_{conc} }{V_{dil}}$

• Now, we can fill the table

Beaker label	Vol prec. Sol	Inital conc NaOH	moles NaOH		Vol H2O used	Vol diluted sol	Final conc NaOH
			Before Dil	After dil
B	35.00 mL	0.012 M	4.20x10-4 mol	4.20x10-4 mol	14.84 mL	35 + 14.84 = 49.84 mL	8.43x10-3 M
C	25.00 mL	8.43x10-3 M	2.11x10-4 mol	2.11x10-4 mol	26.16 mL	25 + 26.16 = 51.16 mL	4.12x10-3 M
D	15.00 mL	4.12x10-3 M	6.18x10-5 mol	6.18x10-5 mol	35.00 mL	15 + 35.00 = 50.00 mL	1.24x10-3 M
E	10.00 mL	1.24x10-3 M	1.24x10-5 mol	1.24x10-5 mol	40.01 mL	10 + 40.01 = 50.01 mL	2.48x10-4 M
F	5.00 mL	2.48x10-4 M	1.24x10-6 mol	1.24x10-6 mol	44.02 mL	5 + 44.02 = 49.02 mL	2.53x10-5 M

Now we can answer the Post-Lab questions:

1) Volume of Sol A (0.012 M) need to prepare 50 mL of sol F (2.53x10^-5 M)

The volume requires of solution A is:

$[NaOH]_{A}\times V_{A} =[NaOH]_{F}\times V_{F}$

$V_{A} =\frac{[NaOH]_{F}\times V_{F} }{[NaOH]_{A}}$

$V_{A} =\frac{2.53x10^{-5}\, M \times 50.00\, mL }{0.012\, M}=\mathbf{0.105\, mL}$

If one drop of a liquid is 0.05 mL, then the needed drops of solution A are:

$drops_{sol\, A} =0.105\, mL\times \frac{1\, drop }{0.05\, mL}=2.11\, drops\approx \mathbf{2\, drops}$

2) Effect of dilution in pH

The dilution of a NaOH solution will decrease the pH of the diluted solution. This is because the pH is related to the concentration of NaOH. The pH could be defined as

$pH=14-pOH=14-(-log\left [OH^{-} \right ])$

where the [OH^-] is equal to NaOH concentration, because all OH^- comes from NaOH. Then if we dilute a NaOH solution, the resulting solution will have a smaller OH^- concentration and a lower pH.

If the dilution process was continued indefinitely. the amount of OH^- from NaOH will be insignificant in comparison with the amount of water. Then, it will be as if the solution contains only water, this means that at infinite dilution the NaOH solution will reach a pH = 7 (the pH of water)