Asked by SteveJanker
In the titration of 45 mL of liquid drain cleaner containing NaOH, 25 mL of 0.75 M HNO3 must be added to reach the equivalence point. What is the molarity of the base in the cleaner?
Answers
Answered by
DrBob222
NaOH + HNO3 ==> NaNO3 + H2O
mols HNO3 = M x L = 0.75 M x 0.045 L = ?
From the equation you know 1 mol NaOH = 1 mol HNO3; therefore,
mols NaOH = mols HNO3 from above.
M NaOH = mols NaOH/L NaOH = ?
Post your work if you get stuck.
mols HNO3 = M x L = 0.75 M x 0.045 L = ?
From the equation you know 1 mol NaOH = 1 mol HNO3; therefore,
mols NaOH = mols HNO3 from above.
M NaOH = mols NaOH/L NaOH = ?
Post your work if you get stuck.
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