<span>What
is the ph of an acetic acid solution if 10 drops are titrated with 70
drops of a 0.65 m koh solution? (ka for acetic acid = 1.8 x 10-5)?
[KOH] = 0.65 M
[OH] = 0.65 M
</span>Dilute your mom
<span>[OH]Dil= 0.65 M * 70/80 = 0.56875 M
pH = 5.4
</span>
Answer: 0.86g/mL
Explanation:
Mass of empty cylinder = 23.731g
Mass of cylinder + liquid = 26.414g
Mass of the liquid = 26.414 — 23.731
= 2.683g
Volume of the liquid = 3.12mL
Density = Mass / volume
Density = 2.683g / 3.12mL
Density = 0.86g/mL
Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
Answer:
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