How many mL of 0.139 M KOH are needed to reach the equivalence point when 1.8 g of potassium hydrogen phthalate, KHP is titrated to a phenolphthalein end-point.

Write the equation and balance it.

Convert 1.8 g KHP to moles. moles = grams/molar mass
Convert moles KHP to moles NaOH using the coefficients in the balanced equation.
Finally, M NaOH = moles NaOH/L NaOH. Solve for L NaOH and convert to mL.

That's KOH and not NaOH.

1.22

To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between potassium hydroxide (KOH) and potassium hydrogen phthalate (KHP).

The balanced equation for the reaction is:
KOH + KHP → K2HP + H2O

1. First, we need to convert the mass of KHP to moles. The molar mass of KHP is 204.22 g/mol, so:

Number of moles of KHP = Mass of KHP / Molar mass of KHP
= 1.8 g / 204.22 g/mol

2. Next, we need to determine the stoichiometric ratio between KOH and KHP from the balanced equation. The equation shows that 1 mole of KOH reacts with 1 mole of KHP. Therefore, the stoichiometric ratio is 1:1.

3. Now, we can calculate the number of moles of KOH required to react with the calculated moles of KHP:

Number of moles of KOH = Number of moles of KHP

4. Finally, to determine the volume of 0.139 M KOH required, we can use the formula:

Volume (in mL) = Number of moles of KOH / Molarity of KOH

Let's substitute the values into the formula to calculate the volume:

Volume of KOH = Number of moles of KOH / Molarity of KOH
= (1.8 g / 204.22 g/mol) / 0.139 mol/L

Now, we need to convert the concentration of KOH from mol/L to moles/mL:

0.139 mol/L = 0.139 mol / 1000 mL
So, the volume of KOH will be:

Volume of KOH = (1.8 g / 204.22 g/mol) / (0.139 mol / 1000 mL)

By evaluating this expression, we can determine the volume of KOH in mL required to reach the equivalence point when 1.8 g of KHP is titrated to a phenolphthalein end-point.