Asked by Jess
A 0.2800 g sample of an unknown acid requires 28.22 ml of 0.1199 M NaOH for neutralization to a phenolphthalein end point.
a.) How many moles of OH- are used?
b.) How many moles of H+ are found in the acid?
c.) What is the calculated equivalent molar mass of the unknown acid?
If I am doing this right, I found a.) to be .003384 mols of OH- by:
28.22 ml of NaOH= .02822 L of NaOH x .1199 M NaOH x (1 mol OH-/1 mol Na+)= .003384 mols OH-. Is that correct? If so, I still have no idea where to start on the following questions. Any guidance would be greatly appreciated.
a.) How many moles of OH- are used?
b.) How many moles of H+ are found in the acid?
c.) What is the calculated equivalent molar mass of the unknown acid?
If I am doing this right, I found a.) to be .003384 mols of OH- by:
28.22 ml of NaOH= .02822 L of NaOH x .1199 M NaOH x (1 mol OH-/1 mol Na+)= .003384 mols OH-. Is that correct? If so, I still have no idea where to start on the following questions. Any guidance would be greatly appreciated.
Answers
Answered by
DrBob222
a. correct
b. mols OH^- = mols H^+
c. mols = grams/molar mass. you have mols and grams, solve for molar mass.
Technically you must know if this is a monoprotic, diprotic or triprotic acid.
b. mols OH^- = mols H^+
c. mols = grams/molar mass. you have mols and grams, solve for molar mass.
Technically you must know if this is a monoprotic, diprotic or triprotic acid.
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