Number of moles of proflavin present in the mixture
=Molarity x Volume (L)
=5.0 x 10-6 M x 0.45 mL/1000 mL/L (1 L=1000 mL)
=2.25 x 10-9 mol
Total volume of solution=0.45 mL + 0.45 mL=0.90 mL
=0.90 mL/1000 mL/L=0.90 x 10-3 L
Concentration of Proflavin in the mixture
=number of moles/volume of solution (L)
=2.25 x 10-9 mol/0.90 x 10-3 L=2.5 x 10-6 M
So the answer is 2.5 x 10-6 M
(3 pts) 6) A solution is prepared by mixing 0.45 mL of 5.0 x 10ⓇM proflavin,...
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A buffer is prepared by mixing 115 mL of a 0.120 M solution of the weak acid HA with 115 mL of a 0.150 M solution of NaA. What will be the pH of the resulting solution? Ka for HA is 2.30x10-6
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