D) blue
(about 400 nanometers)
Complete Question
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?
Answer:
The pK_a value is
Explanation:
From the question we are told
The volume of base is 
The pH of solution is 
The concentration of the acid is 
From the pH we can see that the titration is between a strong base and a weak acid
Let assume that the the volume of acid is 
Generally the concentration of base

Substituting value


When 13mL of the base is added a buffer is formed
The chemical equation of the reaction is

Now before the reaction the number of mole of base is
![No \ of \ moles[N_B] = C_B * V_B](https://tex.z-dn.net/?f=No%20%5C%20of%20%5C%20moles%5BN_B%5D%20%20%3D%20%20C_B%20%2A%20V_B)
Substituting value

Now before the reaction the number of mole of acid is

Substituting value


Now after the reaction the number of moles of base is zero i.e has been used up
this mathematically represented as

The number of moles of acid is


The pH of this reaction can be mathematically represented as
![pH = pK_a + log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%20%3D%20pK_a%20%2B%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
Substituting values

Answer:
2.7724 g
Explanation:
Mass of pre- 1892 pennies = 3.1 g
Abundance = 45.4 %
Mass of post 1892 pennies = 2.5 g
Abundance = 100 - 45.4 = 54.6 %
The average mass is given as = ( 3.1 g * 45.4 / 100) + (2.5g * 54.6 / 100)
Average Mass = 3.1 * 0.454 + 2.5 * 0.546
Average Mass = 1.4074 + 1.365 = 2.7724 g
Remember that a conjugate acid-base pair will differ only by one proton.
None of the options you listed are conjugate acid-base pairs as none of them differ only by one proton (or H⁺)
An example of a conjugate acid-base pair would be NH₃ and NH₄⁺NH₃ + H₂O --> NH₄⁺ + OH⁻NH3 is the base, and NH₄⁺ is the conjugate acid