A 10 milliliter sample of HCl pickle is titrated with 0.2 N NaOH to a phenol-phthalein end point and it is found that 6.5 milliliters are used. What is the concentration of HCl in the Pickle in grams per liter? In weight Percent?
To find the concentration of HCl in the pickle in grams per liter (g/L), we need to use the equation:
Concentration (g/L) = (Volume of NaOH used (L) × Normality of NaOH × Molar mass of HCl) / Volume of pickle sample (L)
First, let's convert the volumes from milliliters (mL) to liters (L):
- Volume of NaOH used = 6.5 mL = 6.5/1000 L = 0.0065 L
- Volume of pickle sample = 10 mL = 10/1000 L = 0.01 L
Since the concentration of NaOH is given as 0.2 N, it means that 1 L of NaOH solution contains 0.2 moles of NaOH. The molar mass of HCl is approximately 36.46 g/mol.
Now we can calculate the concentration of HCl in grams per liter:
Concentration (g/L) = (0.0065 L × 0.2 N × 36.46 g/mol) / 0.01 L
Concentration (g/L) = 4.469 g/L
Therefore, the concentration of HCl in the pickle is approximately 4.469 g/L.
To calculate the concentration in weight percent, we need to know the density of the pickle sample. Let's assume the density is 1 g/mL to simplify the calculation.
Weight percent (%) = (Concentration (g/L) / Density of pickle sample (g/mL)) × 100
Weight percent (%) = (4.469 g/L / 1 g/mL) × 100
Weight percent (%) = 446.9%
Therefore, the concentration of HCl in the pickle is approximately 4.469 g/L or 446.9% (w/w).