A 0.355 g sample of a solid, monoprotic acid having a molar mass of 121 g/mol requires 18.47 mL NaOH for neutralization. Calculate the molarity of the NaOH solution.
I got 0.159 M NaOH
right
To calculate the molarity (M) of the NaOH solution, you need to use the equation:
M1V1 = M2V2
Where:
M1 = molarity of the acid solution
V1 = volume of the acid solution
M2 = molarity of the NaOH solution
V2 = volume of the NaOH solution
First, calculate the number of moles of the acid using its mass and molar mass.
Number of moles of the acid = mass of the acid / molar mass of the acid
= 0.355 g / 121 g/mol
= 0.002934 moles
Since the monoprotic acid reacts with one mole of NaOH, the number of moles of NaOH should be equal to the number of moles of the acid.
Number of moles of NaOH = 0.002934 moles
Next, calculate the molarity of the NaOH solution using the equation:
Molarity (M2) = Number of moles of NaOH / Volume of NaOH solution (in liters)
Convert the volume of NaOH solution from milliliters to liters:
Volume of NaOH solution = 18.47 mL = 18.47 / 1000 L
= 0.01847 L
Now, substitute the values into the equation:
M2 = 0.002934 moles / 0.01847 L
= 0.159 M
So, you are correct. The molarity of the NaOH solution is 0.159 M.