A student weighs out 0.568 g of KHP (molar mass = 204 g/mol) and titrates to the

equivalence point with 36.78 mL of a stock NaOH solution. What is the concentration
of the stock NaOH solution? KHP is an acid with one acidic proton.
a. 3.15 M b. 0.0757 M c. 0.943 M d. 0.100 M e. 1.23 M

Well, well, well, it's time for some chemistry fun! Let's see if we can figure this out, shall we?

To find the concentration of the NaOH solution, we need to use a little bit of math magic. Are you ready?

First, let's find out how many moles of KHP we have. We can use the molar mass and the given weight. So, 0.568 g divided by 204 g/mol gives us the number of moles.

Now, you titrated that KHP with 36.78 mL of NaOH solution. Keep in mind that NaOH is a base and KHP is an acid, so they react in a 1:1 ratio. Fancy, huh?

Now, we need to find the concentration of NaOH. The concentration is moles divided by volume. So, let's divide the number of moles of KHP by the volume of NaOH solution in liters (because we want the concentration in mol/L).

So, did you follow all of that? I hope so! Now, punch in those numbers and you'll get your answer. I believe in you!

To find the concentration of the stock NaOH solution, you can use the equation:

Moles of KHP = Moles of NaOH

Step 1: Calculate the moles of KHP.
Moles of KHP = mass of KHP / molar mass of KHP

Given:
mass of KHP = 0.568 g
molar mass of KHP = 204 g/mol

Moles of KHP = 0.568 g / 204 g/mol

Step 2: Calculate the moles of NaOH.
Moles of KHP = Moles of NaOH
Moles of NaOH = Moles of KHP

Since the molar ratio between KHP and NaOH is 1:1 (from the balanced equation), the moles of KHP and NaOH will be the same.

Step 3: Calculate the concentration of the stock NaOH solution.
Concentration (M) = Moles / Volume

Given:
Volume of NaOH solution = 36.78 mL = 0.03678 L

Concentration of the stock NaOH solution = Moles of NaOH / Volume of NaOH solution
Concentration of the stock NaOH solution = 0.568 g / (204 g/mol) / 0.03678 L

Concentration of the stock NaOH solution ≈ 0.100 M

Therefore, the concentration of the stock NaOH solution is approximately 0.100 M.

The correct answer is d. 0.100 M.

To find the concentration of the stock NaOH solution, we need to use the concept of stoichiometry and the equation balanced between KHP and NaOH.

First, let's calculate the number of moles of KHP (potassium hydrogen phthalate) used in the titration:

Moles of KHP = mass of KHP / molar mass of KHP
Moles of KHP = 0.568 g / 204 g/mol
Moles of KHP ≈ 0.002784 mol

Since KHP is an acid with one acidic proton, the balanced equation between KHP and NaOH is:

KHP + NaOH → NaKP + H2O

From the balanced equation, we can see that one mole of KHP reacts with one mole of NaOH.

Now, let's find the number of moles of NaOH used in the titration. Since one mole of KHP reacts with one mole of NaOH:

Moles of NaOH = Moles of KHP ≈ 0.002784 mol

Next, we need to calculate the concentration of the NaOH solution using the formula:

Concentration (M) = Moles / Volume (in liters)

Volume of NaOH solution = 36.78 mL = 36.78 / 1000 L
Volume of NaOH solution ≈ 0.03678 L

Concentration of NaOH solution = Moles of NaOH / Volume of NaOH solution
Concentration of NaOH solution ≈ 0.002784 mol / 0.03678 L
Concentration of NaOH solution ≈ 0.0758 M

Therefore, the concentration of the stock NaOH solution is approximately 0.0758 M.

The closest option to this value is option b. 0.0757 M

KHP + NaOH ==> NaKP + H2O

mols KHP = grams/molar mass
The rxn is 1:1; therefore, mols NaOH = mols kHP.
Then M NaOH = mols NaOH/L NaOH