A solution of 25 mL of HCL, unknown concentration, is titrated to its equivalence point with 41mL of 1.50 M NaOH. What is the concentration of the acid?
2007-04-25 08:10:34 · 2 answers · asked by cylo1818 3 in Science & Mathematics ➔ Chemistry
You need to use this equation: M1V1=M2V2
M= molarity (moles/L)
V=Volume in L
M1*(0.025L)=(0.041L)*(1.50M)
M1=2.46 M
You have the same number of moles of H+ than of OH- at equilibrium.
2007-04-25 08:17:46 · answer #1 · answered by Anonymous · 1⤊ 0⤋
M.M. got the right answer and got it first, and you should award him/her full points. Another method is the factor-label method, where you start with what's given and work toward the answer.
41mLNaOH/25mLHCl x 1.50molNaOH/1000mLNaOH x 1molHCl/1molNaOH x 1000mLHCl/1LHCl = 2.46molHCl/1LHCl or 2.5M to two significant figures.
The first factor comes from the titration data. The second factor comes from the molarity of NaOH. The mL NaOH cancel leaving mol NaOH. The third factor comes from the balanced equation. The mol NaOH cancel leaving mol HCl. The fourth factor comes from converting mL to L to get mol/L, which is molarity. Writing it out, you see that the 1000's cancel, simplifying the arithmetic.
2007-04-25 15:36:45 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋