I am giving a titration curve of a weak base and strong acid. The pH on the y scale is from 0-14 and the volume of titrant in mL is from 0-34. It looks like the equivalene point is around 22? The question is:

"The above titration curve was obtained when a 10 mL sample of a 0.45 M base was titrated with an acid. What is the approximate molarity of the acid used?" (Assume a 1:1 molar ratio)

You must mean 22 mL?(not pH of 22). Then

mL acid x M acid = mL base x M base

Yes I meant 22mL. How would I go about solving for the Kb?

At the half way point to the equivalence (that would be 22/2 = 11 mL), pH at that point will be pKb

To solve this question, we need to analyze the given titration curve and use the molar ratio between the acid and the base.

First, let's understand the titration curve. The x-axis represents the volume of titrant added (in mL), and the y-axis represents the pH. At the equivalence point, the pH should be around 7, indicating that the acid and base have reacted in stoichiometric proportions.

In this case, the equivalence point is around 22 mL. This means that the volume of the acid added to reach the equivalence point is 22 mL. Now we can use this information to determine the molarity of the acid.

We are given a 10 mL sample of a 0.45 M base. Since the acid and base react in a 1:1 molar ratio, this means that they have equal moles at the equivalence point.

First, we can use the equation:

moles of base = volume of base (in L) * molarity of base

Since we have a 10 mL sample (0.010 L) of the base with a concentration of 0.45 M, we can calculate the moles of base:

moles of base = 0.010 L * 0.45 M = 0.0045 moles

Since the acid and the base react in a 1:1 molar ratio, the moles of acid at the equivalence point is also 0.0045 moles.

Now, to calculate the molarity of the acid, we can use the equation:

molarity of acid = moles of acid / volume of acid (in L)

We know that the volume of acid added at the equivalence point is 22 mL (0.022 L) because the equivalence point is reached at around 22 mL on the titration curve.

molarity of acid = 0.0045 moles / 0.022 L ≈ 0.205 M

Therefore, the approximate molarity of the acid used is 0.205 M.