Hey, lovely! It's a pretty lengthy process but here is a pretty clear video on how to do it. Hope this helps ya!
https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equat...
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
65 grams of HCl = 65/36.5 moles of HCl = 1.78 moles
1.78 moles of HCl dissolved to make a 5 litres of solution has a concentration of 1.78/5 = 0.36 mol/dm^3 (Note: 1 litre = 1 cubic decimetre)
In a strong acid, such as HCl, [H+] = [acid], so [H+] = 0.36
To calculate pH, we have to take the negative logarithm of the concentration of protons
So, -log(0.36) = 0.45
Hope I helped!! xx
NaPO4 + KOH -> KPO4 + NaOH
already balance