A titration of 15.0 cm3 of household ammonia, NH3, required 38.57 cm3 of 0.78M HCl. Calculate the molarity of the ammonia.
please show work and detail.
Worked th same way as the HNO3/KOH problem.
To calculate the molarity of the ammonia (NH3) solution, we can use the equation:
M1V1 = M2V2
where:
M1 = Molarity of HCl solution
V1 = Volume of HCl solution
M2 = Molarity of NH3 solution
V2 = Volume of NH3 solution
In this case, we are given:
M1 = 0.78 M (molarity of HCl solution)
V1 = 38.57 cm3 (volume of HCl solution)
V2 = 15.0 cm3 (volume of NH3 solution)
We need to solve for M2, the molarity of the NH3 solution.
Using the equation M1V1 = M2V2, we can rearrange the equation to solve for M2:
M2 = (M1V1) / V2
Now, let's substitute the values:
M2 = (0.78 M * 38.57 cm3) / 15.0 cm3
Calculating this expression gives:
M2 = 2.01 M
Therefore, the molarity of the ammonia solution (NH3) is 2.01 M.
Note: When interpreting the result, it is important to consider the significant figures. Since the given values have three significant figures, the answer should also be rounded to three significant figures. In this case, the molarity of the ammonia solution would be rounded to 2.01 M.