Login or Sign Up

Ask a New Question

A student finds that 1.00 g of a monoprotic acid, HA , requires 50.0 ml of 0.100 M KOH to neutralize it. What is the molar mass of the acid?

moles HA = moles KOH.

moles KOH = M x L = ??
moles HA = g/molar mass
Solve for molar mass.

Similar Questions

A student finds that 1.00 g of a monoprotic acid, HA , requires 50.0 ml of 0.100 M KOH to neutralize it. What is the molar mass

Top answer: To find the molar mass of the acid (HA), we need to use the information given about its reaction Read more.
Aspirin (acetylsalicylic acid, ) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved

Top answer: It tells you it wants you to calculate the Ka for ASA. If we call it HA, then HA ==> H^+ + A^- Ka = Read more.
A 0.2053g sample of an organic acid requires by titration, 15.0 mL of 0.1008 Msodium hydroxide for complete neutralization.

Top answer: mols NaOH = M x L = ? mols acid = same if it is monoprotic. mols acid = grams/molar mass. You know Read more.
A 0.225g sample of a weak monoprotic acid requires 37.50 mL of 0.100 M NaOH solution to reach the equivalence point. What is the

Top answer: HA + NaOH ==> NaA + H2O moles NaOH added = M x L = ?? moles HA must be the same (see the Read more.
A student is asked to standardize a solution of barium hydroxide. He weighs out 0.986 g potassium hydrogen phthalate (KHC8H4O4,

Top answer: i got it, it's 0.111M! Read more.
A student is asked to standardize a solution of barium hydroxide. He weighs out 0.945 g potassium hydrogen phthalate (KHC8H4O4,

Top answer: These are essentially 4-step problems. 1. Write and balance the equation. If we call potassium Read more.
A 0.040M solution of a monoprotic acid is 14% ionized. Calculate the Ka for the weak acid.---------------- so: HX <---> H+ + X-

Top answer: So (H^+) = 0.04 x 0.14 = ?? (X^-) is the same. (HX) = either, 0.04-(0.04*0.14) OR 0.04*0.86 (HX) Read more.
A 0.649 m aqueous solution of a monoprotic acid (HA) freezes at -2.34°C. Find the pKa of this monoprotic acid. Kf of water =

Top answer: dT = i*Kf*m i = 2.34/1.86*0.649 i = 1.938 so it's not quite 2.0 which means it's 93.8% ionized. Read more.
Calculate the Molar Enthalpy of Neutralization (ÄHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base,

Top answer: To calculate the molar enthalpy of neutralization (ΔHn) in kJ/mol, we need to follow a few steps: Read more.
2) Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic

Top answer: You haven't used specific heat H2O anywhere. q = mass H2O X specific heat H2O x delta T + Ccal*delta Read more.