0.25 g of acid is titrated with .108 M NaOH. if it takes 27.22 mL of NaOH to reach the endpoint, what is the equivalent weight of the acid?
mL x N x milliequivalnt weight = grams
The normality of NaOH is the same as the molarity.
Multiply me wt by 1000 to obtain equivalent weight.
To find the equivalent weight of the acid, we need to first find the number of moles of NaOH used in the titration.
1. Convert the volume of NaOH used from milliliters (mL) to liters (L):
27.22 mL = 27.22/1000 L = 0.02722 L
2. Use the formula C1V1 = C2V2 to calculate the number of moles (n) of NaOH used:
C1V1 = C2V2
(0.108 M)(0.02722 L) = n
Therefore, n = 0.108 x 0.02722 = 0.00294096 moles
3. Since the acid reacts with NaOH in a 1:1 ratio, the number of moles of the acid is also 0.00294096 moles.
4. Now, we can calculate the equivalent weight (EW) of the acid using the formula:
EW = mass (m) / moles (n)
Since we are given the mass of the acid as 0.25 grams:
EW = 0.25 g / 0.00294096 mol
Calculating this division will give us the equivalent weight of the acid.