KH₂PO₄ hydrolyzes as;
H₂PO₄⁻ + H₂O ↔ H₃PO₄ + OH⁻
Let x amount of H₂PO₄⁻ has reacted with water then,
Kb₁ = [H₃PO₄][OH⁻] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.8-x M
Kb₁ = x² / (0.8 - x)
Given Ka₁ = 7.5 x 10⁻³
so Kb₁ = 1 x 10⁻¹⁴ / (7.5 x 10⁻³) = 1.33 x 10⁻¹²
From this information:
1.33 x 10⁻¹² = x² / 0.8
x = [OH⁻] = 1.03 x 10⁻⁶ M
pOH = - log (1.03 x 10⁻⁶) = 5.99
pH = 14 - pOH = 14 - 5.99 = 8.01
Although there isn’t a picture a graph can be misleading when it doesn’t start at zero, it doesn’t give accurate information, it skips too many numbers, the vertical scale is too big or too small. Hope this helps
Answer: N3 H12 P O3
Explanation:
From the question :
N = 31.57% H = 9.10% P = 23.27%
O= 36.06%
Divide each of the element by their respective relative atomic masses.
N = 31.57 / 14 = 2.26
H = 9.10/ 1 = 9.10
P = 23.27 / 31= 0.750
O =36.06 / 16 = 2.25
Divide each answer by the lowest of them all, we then have:
N = 2.26/ 0.750 = Approx = 3
H = 9.10 / 0.750 = Approx = 12
P = 0.750/ 0.750= 1
O = 2.25 / 0.750 = Approx = 3
The empiral formula is
N3 H12 P O3
Answer:
Explanation:
mole of NaOH present = molarity x volume
= 1.0 X 0.05 = 0.05 mole
<em>Recommended mole of HCl </em>= 1.1 x 0.05 = 0.055
<em>Mole of HCl carelessly added by Jacob </em>= 1.1 x 0.04 = 0.044
From the equation of reaction:
HCl + NaOH ----> NaCl + H2O
The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>
The mole of HCl added by Jacob (0.044) is short of the recommended amount (0.055) and also short of the amount required for a neutral endpoint (0.05). <u>This means that the endpoint will have an excess amount of NaOH and as such, basic instead of the desired acidic endpoint.</u>
