There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together. Fourth, find the square root of the result.

The mean for any set of random variables is additive in the sense that

The difference is also additive, so we have 

This means the mean of  is . 

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers  (even when negative), we have

.

So for this difference, we have

.

So the mean and variance are  and , respectively. In addition to that,  is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.

For this problem we need to determine the equation of our standard curve. This can be done by creating a graph from the data points and determining the slope.

Assuming that a sample of zero concentration will also have zero absorbance, we can find the equation for the line generated by finding the slope.

Pick two points on the line to find the slope. We will use (0,0) and (40,0.405).

Use this equation and the absorbance given in the question to find the concentration of the unknown sample.

The slope of the line, , can be calculated by the following equation:

The subscripts are simply the points you chose in the order you arbitrarily chose.The most important thing to do is plug the values into equation in the order you chose them.

To determine the slope of the line we need to use two points on the graph. Assuming the two points chosen correspond to  and , plugging these values into the equation gives:

Therefore the slope of the line is equal to 

For a general chemical equation , where A, B, C, and D are elements and the Greek letters are their coefficients, we have the reaction quotient equation:

We can find the reaction quotient equation for our reaction by substituting the variables.

Notice that the concentration of  is in the denominator and is squared, so doubling the concentration of  changes the reaction quotient by a factor of one-fourth.

For any titration curve the equivalence point corresponds to the steepest part of the curve. In this case, the question indicates that the pH at equivalence was 8.9, which would correspond to a volume between 20.0 and 24.0 on the x-axis of the graph. This leads to our approximate answer of 21.8mL.

The pK_a corresponds to the pH when half of the acid has been neutralized. In this case, that would correspond to when half of the 0.5M NaOH has been added. We can see in the curve that the equivalence point is at around 22mL of NaOH, as this is the steepest change in pH. The half equivalence point will be found at approximately 11mL of NaOH addition. On the graph, 11mL of NaOH results in a pH of approximately 5.0, giving us our pK_a value.

H₂CO₃ will have two equivalence points and two half equivalence points, one set corresponding to each of its two protons.

After the first equivalence point, all of the acid is in the form HCO₃^-. When an acid is at a half equivalence point, the acid's concentration will be equal to the concentration of the conjugate base. For the second half equivalence point, the acid is in the  HCO₃^- form, and the conjugate base is CO₃^2-. As a result, at the second half equivalence point, the concentrations of HCO₃^- and CO₃^2- will be equal.

To solve this question you need to think about the chemical reaction occurring.

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pK_a when the concentration of conjugate base (acetate) equals the concentration of acid.

If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.

L = mol/M = (0.5mol)/(2M) = 0.25L

To answer this question, we need to be able to compare the concentrations of acid and conjugate base in the solution with the pH. The Henderson-Hasselbach equation is used to compare these values, and is written as:. In this question, this equation can be written with the given values of K_a and pH.

Since we know the K_a of HCN, we can derive the pK_a, which turns out to be 9.2.

As a result, we want to see to it that the amount of conjugate base is equal to the concentration of acid, so that . Because log(1) = 0, we want to see to it that the concentrations of the acid and the conjugate base are equal to one another. We know from the question that [HCN] = 2M.

As a result, a concentration of 2M NaCN will allow the pH of the solution to be 9.2.