What will be the molarity of the resulting solutions made by mixing the following? Assume that volumes are additive.
91 mL of 0.42 M HNO3 with 59 mL of 1.6 M HNO3
i know that i'm going to use the formula M1V1=M2V2, but i'm not sure what numbers to substitute to where.
To determine the molarity of the resulting solution when two solutions are mixed, you can use the equation:
(M1 × V1) + (M2 × V2) = (M3 × V3)
Where:
M1 = molarity of the first solution
V1 = volume of the first solution
M2 = molarity of the second solution
V2 = volume of the second solution
M3 = molarity of the resulting solution (what you're trying to find)
V3 = volume of the resulting solution
In this case, you want to find the molarity of the resulting solution when 91 mL of 0.42 M HNO3 is mixed with 59 mL of 1.6 M HNO3.
So, let's substitute the known values into the equation:
(M1 × V1) + (M2 × V2) = (M3 × V3)
(0.42 M × 91 mL) + (1.6 M × 59 mL) = (M3 × (91 mL + 59 mL))
Now, we can calculate the value:
M3 × (91 mL + 59 mL) = (0.42 M × 91 mL) + (1.6 M × 59 mL)
M3 × 150 mL = (0.42 M × 91 mL) + (1.6 M × 59 mL)
Now, we substitute the values and solve for M3:
M3 × 150 mL = 38.22 Mm + 94.4 Mm
M3 × 150 mL = 132.62 Mm
M3 = 132.62 Mm / 150 mL
After performing the calculation, you will find the molarity of the resulting solution (M3) by dividing 132.62 Mm by 150 mL.
Therefore, the molarity of the resulting solution can be determined by dividing 132.62 Mm by 150 mL.