determine the approximate molarity of a solution of NaOH prepared by diluting 10ml of 6M NaOH to a volume of 400 ml.
using the solution in the first part, 24.5ml of NaOH solution was required to neutralize 26.4ml of a .150M HCl solution. find the molarity of NaOH
good luck chuck!
Approx molarity = 6 M x (10/400) or about 0.15 M
exact molarity is
M x mL = M x mL
?M x 24.5 mL = 0.15 x 26.4 mL
solve for ?M.
0.162
To determine the molarity of the NaOH solution, we can use the concept of dilution. The molarity of a solution is given by the formula:
M1V1 = M2V2
where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
In the first part of the question, 10 ml of a 6M NaOH solution is diluted to a final volume of 400 ml. We can use this information to find the final molarity.
M1 = 6M
V1 = 10 ml
V2 = 400 ml
M2 = (M1 * V1) / V2
= (6M * 10 ml) / 400 ml
= 0.15 M
So, the molarity of the NaOH solution prepared by diluting 10 ml of 6M NaOH to a volume of 400 ml is approximately 0.15 M.
In the second part of the question, we are given that 24.5 ml of the NaOH solution from the first part is required to neutralize 26.4 ml of a 0.150 M HCl solution. We can use this information to find the molarity of the NaOH solution.
M1 = 0.15 M
V1 = 24.5 ml
M2 = 0.150 M
V2 = 26.4 ml
Now, we can rearrange the formula to solve for M1:
M1 = (M2 * V2) / V1
= (0.150 M * 26.4 ml) / 24.5 ml
≈ 0.162 M
So, the molarity of the NaOH solution is approximately 0.162 M.