The pH or acidity of liquid is measured in pH, where pH is given by the equation pH=- log C, with C being the concentration of [H+] in multiples of M=1mol/L. Suppose you are given a solution of hydrochloric acid with a pH of 1.7 and asked to increase the pH solution by 1.4. Determine by how much you must dilute the solution if the ratio of the concentration of two acids is c2/c1 . Does your answer differ if you start with pH of 2.2

To determine how much you must dilute the hydrochloric acid solution, we need to calculate the change in concentration required.

Given that pH is calculated using the equation pH = -log[C], we can rearrange it to find the concentration [C]:

[C] = 10^(-pH)

Let's calculate the concentration of the initial solution with a pH of 1.7:

[C1] = 10^(-1.7) = 0.01995 M

Now, we want to increase the pH by 1.4 units. To do this, we need to calculate the concentration of the new solution:

[C2] = 10^(- (1.7 + 1.4))
[C2] = 10^(-3.1) = 0.0007943 M

To determine the dilution factor, we need to find the ratio of the concentrations of the two solutions:

c2/c1 = [C2] / [C1]
c2/c1 = (0.0007943 M) / (0.01995 M)
c2/c1 ≈ 0.0398

Therefore, the ratio of the concentrations is approximately 0.0398. This implies that you must dilute the solution by a factor of approximately 0.0398 to achieve the desired pH increase.

If we start with a pH of 2.2, the process is similar. Let's calculate the concentration of the initial solution with a pH of 2.2:

[C1] = 10^(-2.2) = 0.0063096 M

To increase the pH by 1.4 units, we calculate the concentration of the new solution:

[C2] = 10^(- (2.2 + 1.4))
[C2] = 10^(-3.6) = 0.00025 M

Now, let's calculate the ratio of the concentrations:

c2/c1 = [C2] / [C1]
c2/c1 = (0.00025 M) / (0.0063096 M)
c2/c1 ≈ 0.0397

The ratio of the concentrations is approximately 0.0397, which is very close to the previous case. So, the answer does not differ significantly when starting with a pH of 2.2.