The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2
I don’t know what the question is asking
Answer:
0.286 moles
Explanation:
i hope this is helpful for you
N=N₀*2^(-t/T)
N₀=200 g
T=10 d
t=30 d
N=200*2^(-30/10)=25 g
25 g will remain
Answer:
1. 4
2. 2
3. 3
4. 1
Explanation: just did it on edge 2021
