Asked by ag
An unknown amount of water is mixed with 310 mL of a 6 M solution of NaOH solution. A 75 mL sample of the resulting solution is titrated to neutrality with 58.2 mL of 6 M HCl. Calculate the concentration of the diluted NaOH solution. Answer in units of M
What volume of water was added to the 310 mL of NaOH solution? Assume volumes are additive. Answer in units of mL
What volume of water was added to the 310 mL of NaOH solution? Assume volumes are additive. Answer in units of mL
Answers
Answered by
DrBob222
M of the 75 mL =
58.2 mL x 6M/75 mL = 4.656M
So the initial solution must have been
6M NaOH x 310 mL = 4.656M x ? mL
Solve for ? mL, then subract 310 from it to see how much water was added.
58.2 mL x 6M/75 mL = 4.656M
So the initial solution must have been
6M NaOH x 310 mL = 4.656M x ? mL
Solve for ? mL, then subract 310 from it to see how much water was added.