Gold(ill) hydroxide is used in medicine, porcelain making, and gold plating. It is quite insoluble in aqueous solution. Which of
1 answer:
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Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
Answer:
9.25
Explanation:
Let first find the moles of and
number of moles of = 0.40 mol/L × 200 × 10⁻³L
= 0.08 mole
number of moles of = 0.80 mol/L × 50 × 10⁻³L
= 0.04 mole
The equation for the reaction is expressed as:
The ICE Table is shown below as follows:
Initial (M) 0.08 0.04 0
Change (M) - 0.04 -0.04 + 0.04
Equilibrium (M) 0.04 0 0.04
for buffer solutions
since they are in the same solution