Asked by Sam
A 16.6 mL portion of an HBr solution of unknown strength is diluted to exactly 247 mL. If 102.1 mL of this diluted solution requires 87.8 mL of 0.37 M NaOH to achieve complete neutralization, what was the strength of the original HBr solution?
Answers
Answered by
DrBob222
NaOH + HBr ==> NaBr + H2O
moles NaOH used in the titration = M x L = 0.37 x 0.0878 = 0.03249.
Using the balanced equation above, convert moles NaOH to moles HBr. The coefficients are 1:1; therefore, moles HBr = 0.03249
That = the moles in the 102.1 mL sample. How much was in the 247 mL. That will be
0.03249 x (247/102.1) = ?? moles HBr. That = the number of moles HBr in the 16.6 mL original sample.
Then M = moles/L soln.
moles NaOH used in the titration = M x L = 0.37 x 0.0878 = 0.03249.
Using the balanced equation above, convert moles NaOH to moles HBr. The coefficients are 1:1; therefore, moles HBr = 0.03249
That = the moles in the 102.1 mL sample. How much was in the 247 mL. That will be
0.03249 x (247/102.1) = ?? moles HBr. That = the number of moles HBr in the 16.6 mL original sample.
Then M = moles/L soln.
Answered by
Sam
just for clarification,it's
?? moles HBr/original sample (which is 16.6ml)
right?
?? moles HBr/original sample (which is 16.6ml)
right?
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