Asked by Maya
A sample of 1.018 g of KHP (potassium hydrogen phthalate, molar mass = 204.22 g/mol) was dissolved in ~ 25 mL distilled water and titrated with a NaOH solution of unknown concentration.
If 28.69 mL of base was used to reach the endpoint, what was the concentration of the NaOH (in mol/L)?
If 28.69 mL of base was used to reach the endpoint, what was the concentration of the NaOH (in mol/L)?
Answers
Answered by
DrBob222
mool KHP = grams/molar mass
mols NaOH = mols KHP (from the coefficients in the balanced equation.)
M NaOH = mols NaOH/L NaOH
mols NaOH = mols KHP (from the coefficients in the balanced equation.)
M NaOH = mols NaOH/L NaOH
Answered by
Maya
So does [mols NaOH = mols KHP] mean that the values of one is equal to the other?
Answered by
DrBob222
yes.
Write the equation and look at the coefficients.
KHP + NaOH ==> NaKP + H2O
1 mol KHP = 1 mol NaOH
Write the equation and look at the coefficients.
KHP + NaOH ==> NaKP + H2O
1 mol KHP = 1 mol NaOH
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