You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:
PH = Pka + log[acid/base]
Step 1 (Figure out the concentrations):
0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3
Step 2 (Plug into equation):
PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]
PH = 3.69
Answer:
Is this the full question
Explanation:
Reply or love the message so i can get a notification
Answer:
answer: they seek to produce verficiable data.
C) Calcium chloride has stronger inter molecular forces than water.
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.