The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M

calculation

molarity =moles of solute/ Kg of the solvent

mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles

calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol

mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg

molarity is therefore = 0.06/0.09842 = 0.6 M

3 0

3 years ago

An aqueous solution contains 0.29 M of benzoic acid (HA) and 0.16 M of sodium benzoate (A-). If the pH of this solution was meas

Inga [223]

Answer:

pKa = 4.89.

Explanation:

We can solve this problem by using the Henderson-Hasselbach equation, which states:

pH = pKa + log $\frac{[A^-]}{[HA]}$

In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

We input the given data:

4.63 = pKa + log $\frac{0.16}{0.29}$

And solve for pKa:

pKa = 4.89

3 0

2 years ago