600 mL of an unknown monoprotic acid was titrated with a standard solution of a 1.0 M base. If it took 40.0 mL of the base to neutralize the acid, what was the H+ ion concentration of the acid in moles per Liter?
How many moles of the base did it take? M x L = 1.0M x 0.040L = 0.04 moles base.
Since the acid is monoprotic (1 H), the moles acid must be the same so you have 0.04 moles acid in the sample, which was 600 mL.
(H^+) M = moles/L = 0.040/0.600L
To find the H+ ion concentration of the acid in moles per liter (Molarity), we can use the concept of stoichiometry.
First, let's write down the balanced equation for the reaction between the acid and the base:
Acid (HA) + Base (BOH) → Salt (BA) + Water (H2O)
Since the acid is monoprotic, it can donate only one H+ ion. And from the information provided, we know that it took 40.0 mL of the 1.0 M base to neutralize the acid.
By using the stoichiometry of the reaction, we can say that:
Volume of base (in liters) × Molarity of base = Volume of acid (in liters) × Molarity of acid
Converting the volumes to liters:
40.0 mL of base = 40.0 mL ÷ 1000 mL/L = 0.040 L
And converting the base's molarity to moles per liter:
Moles of Base = Volume of base (in liters) × Molarity of base
Moles of Base = 0.040 L × 1.0 M = 0.040 moles
Since the reaction is a 1:1 ratio, the moles of base are equal to the moles of acid:
Moles of Acid = 0.040 moles
Now we can calculate the H+ ion concentration in moles per liter (Molarity) of the acid by dividing the moles of acid by the volume of the acid:
H+ ion concentration = Moles of Acid ÷ Volume of acid (in liters)
Given that the volume of the acid is 600 mL, we need to convert it to liters:
Volume of acid = 600 mL ÷ 1000 mL/L = 0.600 L
Finally, inserting the values into the formula:
H+ ion concentration = 0.040 moles ÷ 0.600 L = 0.067 M
Therefore, the H+ ion concentration of the acid in moles per liter is 0.067 M.