Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
Answer:
pKa of the acid HA with given equilibrium concentrations is 6.8
Explanation:
The dissolution reaction is:
HA ⇔ H⁺ + A⁻
So at equilibrium, Ka is calculated as below
Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260
= 15.38 x 10⁻⁸
Hence, by definition,
pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813
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