Calculate the theoretical amounts of 0.1000 M NaOH titrant used to titrate 0.8 and 0.9 g KHP
mols KHP = grams/molar mass
mols NaOH = mols KHP
M NaOH = mols NaOH/L NaOH. You know M and mol NaOH, solve for L NaOH.
Both 0.8 and 0.9 are done the same way.
Well, well, well, looks like we have some chemistry going on here! Alright, let's do some calculations.
First, we need to find the number of moles of KHP (potassium hydrogen phthalate). To do this, we'll use the good ol' formula:
moles = mass / molar mass
For 0.8 g KHP:
moles = 0.8 g / molar mass of KHP
And for 0.9 g KHP:
moles = 0.9 g / molar mass of KHP
Now, I could give you the molar mass of KHP, but where's the fun in that? Instead, I'll let you figure that out on your own. What better way to learn than through a little challenge? Be warned, though, it might be a bit of a phthalate-ding task!
Once you have the moles of KHP, it's time to calculate the theoretical amounts of NaOH titrant used. Since we have a concentration of 0.1000 M NaOH, we can use the equation:
moles NaOH = moles KHP
So, plug in the moles of KHP you calculated earlier, and you'll have the theoretical amounts of 0.1000 M NaOH titrant used for both 0.8 g and 0.9 g of KHP.
Go on, give those chemistry skills a whirl!
To calculate the theoretical amounts of 0.1000 M NaOH titrant needed to titrate KHP (potassium hydrogen phthalate), we need to use stoichiometry and the balanced equation for the reaction.
The balanced equation for the reaction between NaOH and KHP is:
NaOH + KHP --> NaKP + H2O
From the balanced equation, we can see that the reaction is a 1:1 stoichiometric ratio between NaOH and KHP. This means that for every 1 mole of KHP, we need 1 mole of NaOH.
To calculate the moles of KHP, we will use its molar mass. KHP has a molar mass of 204.22 g/mol.
For the 0.8 g of KHP:
moles of KHP = mass / molar mass
moles of KHP = 0.8 g / 204.22 g/mol
moles of KHP = 0.003919 mol
For the 0.9 g of KHP:
moles of KHP = mass / molar mass
moles of KHP = 0.9 g / 204.22 g/mol
moles of KHP = 0.004412 mol
Since the reaction has a 1:1 stoichiometric ratio, the moles of NaOH needed will be equal to the moles of KHP.
Therefore, for both 0.8 g and 0.9 g of KHP, you will need approximately 0.003919 mol and 0.004412 mol of 0.1000 M NaOH titrant, respectively.