When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :
when Pka = - ㏒ Ka
6.86 = -㏒ Ka
∴Ka = 1.38 x 10^-7
by using ICE table:
H2PO4- → H+ + HPO4
initial 0.4 m 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka = [H+][HPO4] / [H2PO4-]
by substitution:
1.38 X 10^-7 = X^2 / (0.4-X) by solving for X
∴X = 2.3x 10^-4
∴[H+] = X = 2.3 x 10^-4
∴PH = -㏒[H+]
= -㏒ (2.3 x 10^-4)
∴PH = 3.6
Answer: 121.7558 amu
Explanation:
Average atomic mass of the unknown element =
(Mass of isotope 1 x Relative Abundance of Isotope 1) + ( Mass of Isotope 2 x Relative Abundance of Isotope 2)
(120.9 x 0.5721) + (122.9 x 0.4279) = 69.16689 + 52.58891
Average mass of the unknown element = 121.7558 amu
The limiting reagent is <u>H₂SO₄</u>
<u><em>calculation</em></u>
<u><em> </em></u>Step 1 :write the equation for reaction
2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
Step 2: use the mole ratio to determine the moles of product produced from each reactant
that is from equation above,
NaOH : Na₂SO₄ is 2 :1 therefore the moles of Na₂So₄
= 10.0 moles x 1/2 = 5.0 moles
H₂SO₄ :Na₂SO₄ is 1:1 therefore the moles of Na₂SO₄ is also = 3.50 moles
H₂SO₄ is the limiting reagent since it produces less amount of Na₂SO₄
