7.50gram quantity of a diprotic acid was dissolved in water and made up to exactly 250ml. If 25ml of this solution were neutralized by 11.1 mL of 1.0 M KOH then what is the molar mass of the acid?
I did 7.50/5.55x10^-3
OK EXCEPT that the mass is not 7.5 grams. You titrated only 25 mL aliquot of the original sample of 250 mL; therefore, you titrated only 0.75 g.
Thus 0.75/5.55 x 10^-3 = ?? molar mass
how did you get the 5.55 x 10^-3 ?
.0111/2= .00555
To find the molar mass of the acid, let's break down the problem step by step:
1. Calculate the number of moles of KOH used:
- We are given that 11.1 mL of 1.0 M KOH was used.
- 11.1 mL represents 0.0111 L (since 1 mL = 0.001 L).
- To calculate the number of moles, we use the formula: moles = volume (in liters) × concentration.
- So, moles of KOH = 0.0111 L × 1.0 mol/L = 0.0111 mol.
2. Determine the number of moles of acid neutralized:
- Since KOH is a strong base and diprotic acid means each molecule of acid neutralizes 2 moles of KOH, the number of moles of acid neutralized is half the number of moles of KOH.
- Moles of acid = 0.0111 mol / 2 = 0.00555 mol.
3. Calculate the molar mass of the acid:
- We are given that 7.50 grams of the acid was dissolved in 250 mL (0.250 L) of solution.
- To calculate the molar mass, we use the formula: molar mass = mass (in grams) / moles.
- So, molar mass of the acid = 7.50 g / 0.00555 mol ≈ 1350.67 g/mol.
Therefore, the molar mass of the diprotic acid is approximately 1350.67 g/mol.