Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:
The mixture would contain
if 
undergoes no hydrolysis; the solution is of volume 
after the mixing. The two species would thus be of concentration 
and 
, respectively.
Construct a RICE table for the hydrolysis of under a basic aqueous environment (with a negligible hydronium concentration.)
The question supplied the <em>acid</em> dissociation constant 
for acetic acid 
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant 
for its conjugate base, . The following relationship relates the two quantities:
... where the water self-ionization constant 
under standard conditions. Thus 
. By the definition of :
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)
![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)
Answer:
We could do two 1:50 dilutions and one 1:4 dilutions.
Explanation:
Hi there!
A solution that is 1000 ug/ ml (or 1000 mg / l) is 1000 ppm.
Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.
Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.
We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):
First step (1:50 dilution):
Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).
Step 2 (1:50 dilution):
Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)
Step 3 (1:4 dilution):
Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.
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