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
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
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:
76.5g KCl/74.55 grams per mole Kcl = x
molality= x/.085 kg H2O
Explanation:
well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.
Convert 72g of water into moles of water using molecular weights.
So water is H2O so add up those molecular weights (H=1 and O=16)
2(1)+(16) = 18 g/mol
Then convert so 72g / (18 g/mol) = 4 mol
Now you can convert mol of water to mol of oxygen. So 4 mol of water is 4 mol of oxygen. Then use oxygen molecular weight to find grams again.
4 mol oxygen * 16 g/mol = 64g of oxygen
If we were doing hydrogen instead of oxygen there would be 8 mol hydrogen in 4 mol of water (2 H’s in every H2O molecule) and since we have 74 grams and oxygen is 64 grams, Hydrogen should be 8 grams. Math to check below
8 mol hydrogen * 1 g/mol = 8g of hydrogen
It all adds to 72 so we are correct.
