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

Answered way way above.

To calculate the theoretical amounts of 0.1000 M NaOH titrant used to titrate 0.8 and 0.9 g of KHP (potassium hydrogen phthalate), we need to first determine the molar mass of KHP.

The molar mass of KHP can be calculated by adding the atomic masses of its constituent elements: potassium (K), hydrogen (H), carbon (C), and oxygen (O).

The atomic masses are as follows:
- Potassium (K): 39.10 g/mol
- Hydrogen (H): 1.01 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol

Now we can calculate the molar mass of KHP:
Molar Mass of KHP = 1(39.10 g/mol) + 4(1.01 g/mol) + 8(12.01 g/mol) + 4(16.00 g/mol)
= 39.10 g/mol + 4.04 g/mol + 96.08 g/mol + 64.00 g/mol
= 203.22 g/mol.

Since we know the molar mass of KHP, we can calculate the number of moles of KHP for each sample (0.8 g and 0.9 g).

Number of moles of KHP = Mass of KHP / Molar Mass of KHP

For 0.8 g KHP:
Number of moles of KHP = 0.8 g / 203.22 g/mol
= 0.003935 mol.

For 0.9 g KHP:
Number of moles of KHP = 0.9 g / 203.22 g/mol
= 0.004433 mol.

Now to find the theoretical amount of 0.1000 M NaOH titrant used, we can use the stoichiometry of the reaction between NaOH and KHP.

The balanced chemical equation for the reaction is:
NaOH + KHP -> NaKP + H2O

From the equation, we can see that the stoichiometric ratio between NaOH and KHP is 1:1. This means that 1 mole of NaOH will react with 1 mole of KHP.

So, the theoretical amounts of 0.1000 M NaOH titrant used to titrate 0.8 g and 0.9 g KHP are the same as the number of moles of KHP we calculated earlier.

For 0.8 g KHP: Theoretical amount of NaOH = 0.003935 mol.
For 0.9 g KHP: Theoretical amount of NaOH = 0.004433 mol.