Answer:
1s2 2s2 2p6 3s2 3p6 3d6 4s2
Explanation:
Answer:J , the answer is super giants
First, we need to get moles of NaOH:
when moles NaOH = volume * molarity
= 0.02573L * 0.11 M
= 0.0028 moles
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
= 9.4 x 10^-4 / 0.034 L
= 0.028 M
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
