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
Answer:
b) 5.87 E23 molecules
Explanation:
∴ mm SO3 = 80.066 g/mol
⇒ molecules SO3 = (78.0 g)(mol/80.066 g)(6.022 E23 molec/mol)
⇒ molec SO3 = 5.866 E23 molecules SO3
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
