Calculate the theoretical amounts of 0.1000 M NaOH titrant used to titrate 0.8 and 0.9 g KHP

Please show step by step solution , i need the correct significant figures and units to finish up the rest of the lab correctly i found the answer but for some reason ,using my answers taking me to wrong solution at the end of the lab ?

I did this for you yesterday, in detail.

KHP + NaOH ==> KNaP + H2O
mols KHP = grams/molar mass = 0.8/molar mass KHP = approx 0.0039 but you need to do this step as all that follow more accurately as I've estimated.
Step 2. Convert mols KHP to mols NaOH using the coefficients in the balanced equation. That's approx 0.0039 x (1 mol NaOH/1 mol KHP) = 0.0039 x 1 = approx 0.0030

Step 3. Then M NaOH = mols NaOH/L NaOH.
You know M NaOH from the problem as 0.1000M and you know mols NaOH from step 2, substitute and solve for L NaOH.
Convert to mL if desired.

NOTE. If you already have the answer and are having trouble, the EASY way to solve your problem is to post your work and let us find the error. Telling us you can't get the units or whatever is like fishing in the dark. It's a hit and miss for us. Also, I suggest you type the problem as is; e.g. if that is 0.8g KHP or 0.8000 g KHP make sure it is correct.

Step 3. Its easier to just directly divide the found moles of NaOH from your initial concentration (0.1000 M NaOH). The final answer with correct SF is = 0.04 L or 40 mL

Alright, buckle up for some titration fun!

Step 1: Convert the mass of KHP to moles.

For 0.8 g of KHP:
Molar mass of KHP = (39.10 g/mol K) + (1.01 g/mol H) + (39.10 g/mol P) + (16.00 g/mol O) = 204.22 g/mol

Moles of KHP = (0.8 g KHP) / (204.22 g/mol) = 0.0039201 mol KHP

Rounding to the correct number of significant figures, we get 0.0039 mol KHP.

For 0.9 g of KHP:
Molar mass of KHP = 204.22 g/mol

Moles of KHP = (0.9 g KHP) / (204.22 g/mol) = 0.004408 mol KHP

Again, rounding to the correct number of significant figures, we get 0.0044 mol KHP.

Step 2: Calculate the stoichiometry between NaOH and KHP.

The balanced equation for the reaction between NaOH and KHP is:

NaOH + KHP -> NaKP + H2O

From the equation, we see that the stoichiometric ratio between NaOH and KHP is 1:1. That means for every 1 mol of KHP, we need 1 mol of NaOH.

So, for 0.0039 mol KHP, we need 0.0039 mol of NaOH.

And for 0.0044 mol KHP, we need 0.0044 mol of NaOH.

Step 3: Convert moles of NaOH to volume using the molarity.

We know that the molarity (M) of NaOH is 0.1000 M, which means that we have 0.1000 mol of NaOH per liter of solution.

For 0.0039 mol of NaOH:
Volume of NaOH = (0.0039 mol NaOH) / (0.1000 mol/L) = 0.039 L

And for 0.0044 mol of NaOH:
Volume of NaOH = (0.0044 mol NaOH) / (0.1000 mol/L) = 0.044 L

Now we need to convert liters to milliliters, so multiply each result by 1000:

For 0.0039 mol KHP, we need 39.0 mL of NaOH.

For 0.0044 mol KHP, we need 44.0 mL of NaOH.

There you have it! The theoretical amounts of 0.1000 M NaOH titrant required to titrate 0.8 g and 0.9 g of KHP are 39.0 mL and 44.0 mL, respectively. Remember to use the correct significant figures and units in your calculations to get the correct results in your lab. Good luck!

To calculate the theoretical amounts of 0.1000 M NaOH titrant used to titrate the given amounts of KHP (potassium hydrogen phthalate), follow these steps:

Step 1: Determine the molar mass of KHP.
The molar mass of KHP can be calculated by adding the atomic masses of each element in its chemical formula (C8H5O4):
Molar mass of KHP = (8 x atomic mass of C) + (5 x atomic mass of H) + (4 x atomic mass of O)

Step 2: Convert the given masses of KHP to moles.
To convert the given masses of KHP to moles, divide each mass by the molar mass of KHP:
Moles of KHP = Mass of KHP / Molar mass of KHP

Step 3: Calculate the moles of NaOH required to react with KHP.
The balanced chemical equation between NaOH and KHP is:
NaOH + KHP → NaKP + H2O
From the balanced equation, it can be seen that the stoichiometric ratio of NaOH to KHP is 1:1. This means that 1 mole of NaOH reacts with 1 mole of KHP. Therefore, the moles of NaOH required to react with KHP is equal to the moles of KHP.

Step 4: Convert the moles of NaOH to volume.
Using the given concentration of NaOH (0.1000 M), the moles of NaOH can be converted to volume using the formula:
Volume of NaOH (L) = Moles of NaOH / Molarity (M)

Now, let's perform the calculations for the given masses of KHP: 0.8 g and 0.9 g.

Step 1: Determine the molar mass of KHP.
The molar mass of KHP is 204.23 g/mol.

Step 2: Convert the given masses of KHP to moles.
For 0.8 g of KHP:
Moles of KHP = 0.8 g / 204.23 g/mol
Moles of KHP = 0.00392 mol (rounded to 4 significant figures)

For 0.9 g of KHP:
Moles of KHP = 0.9 g / 204.23 g/mol
Moles of KHP = 0.00441 mol (rounded to 4 significant figures)

Step 3: Calculate the moles of NaOH required to react with KHP.
The moles of NaOH required is equal to the moles of KHP, which we calculated in step 2.

For 0.8 g of KHP:
Moles of NaOH = 0.00392 mol (rounded to 4 significant figures)

For 0.9 g of KHP:
Moles of NaOH = 0.00441 mol (rounded to 4 significant figures)

Step 4: Convert the moles of NaOH to volume.
Using the molarity of NaOH (0.1000 M), we can calculate the volume of NaOH required.

For 0.8 g of KHP:
Volume of NaOH = 0.00392 mol / 0.1000 M
Volume of NaOH = 0.0392 L or 39.2 mL (rounded to 3 significant figures)

For 0.9 g of KHP:
Volume of NaOH = 0.00441 mol / 0.1000 M
Volume of NaOH = 0.0441 L or 44.1 mL (rounded to 3 significant figures)

Therefore, the theoretical amounts of 0.1000 M NaOH titrant required to titrate 0.8 g and 0.9 g of KHP are:
- 0.8 g KHP requires approximately 39.2 mL of 0.1000 M NaOH titrant.
- 0.9 g KHP requires approximately 44.1 mL of 0.1000 M NaOH titrant.