Complete Question
A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane
The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?
Answer:
The mass of caffeine extracted is 
Explanation:
From the question above we are told that
The K value is 
The mass of the caffeine is 
The volume of water is 
The volume of caffeine is 
The number of times the extraction was done is n = 3
Generally the mass of caffeine that will be extracted is
![P = m * [\frac{V}{K * v_c + V} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%20m%20%20%2A%20%20%5B%5Cfrac%7BV%7D%7BK%20%2A%20%20v_c%20%2B%20V%7D%20%5D%5E3)
substituting values
![P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%2040%20%20%20%2A%20%20%5B%5Cfrac%7B2%7D%7B4.6%20%2A%20%202%20%2B%202%7D%20%5D%5E3)

Answer:
1.26 × 10^-8 M
Explanation:
We are given;
Number of moles of mercury (i) chloride as 0.000126 μmol
Volume is 100 mL
We are required to calculate the concentration of the solution.
We need to know that;
Concentration is also known as molarity is given by;
Molarity = Number of moles ÷ Volume
Number of moles = 1.26 × 10^-10 Moles
Volume = 0.01 L
Therefore;
Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L
= 1.26 × 10^-8 M
Thus, the molarity of the solution is 1.26 × 10^-8 M
B
mass of solute - 4.0 g
mass of solution - 100g + 4.0g = 104g
4/104 = 0.03846
0.03846 • 100 = 3.8%
Answer:
A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.
Explanation:
A buffer is a solution which resists changes to its pH when a small amount of acid or base is added to it.
Buffers consist of a weak acid (HA) and its conjugate base (A–) or a weak base and its conjugate acid. Weak acids and bases do not completely dissociate in water, and instead exist in solution as an equilibrium of dissociated and undissociated species. When a small quantity of a strong acid is added to a buffer solution, the conjugate base, A-, reacts with the hydrogen ions from the added acid to form the weak acid and a salt thereby removing the extra hydrogen ions from the solution and keeping the pH of the solution fairly constant. On the other hand, if a small quantity of a strong base is added to the buffer solution, the weak acid dissociates further to release hydrogen ions which then react with the hydroxide ions of the added base to form water and the conjugate base.
For example, if a small amount of strong acid is added to a buffer solution that is 0.700 M H3PO4 and 0.700 M KH2PO4, the following reaction is obtained:
KH₂PO₄ + H+ ----> K+ + H₃PO₄
Therefore, [H₃PO₄] will increase, [KH₂PO₄] will decrease, and pH will slightly decrease.: