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?
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.
A 20.00 ml stock sample of HBr is diluted to 50 mL. If 18.76 ml of 0.1345 M KOH was required to reach the equivalence point, what is the concentration of the stock HBr solution? Is my answer correct? Titration problems can be