Answer:
a) Addition reaction, is your answer
Answer:
b) 3.10
Explanation:
HF ⇄ H
+ + F
Using Henderson-Hasselbalch Equation:
pH = pKa + log [A-]/[HA].
Where;
pKa = Dissociation constant = -log Ka
Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266
[A-] = concentration of conjugate base after dissociation = moles of base/total volume
= 0.15 x 0.3/0.8
= 0.05625 M
[HA] = concentration of the acid = moles of acid/total volume
= 0.10 x 0.5/0.8
= 0.0625 M
Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>
pH = 3.14266 + log [0.05625/0.0625]
= 3.14267 + (-0.04575749056)
= 3.09691250944
<em>From all the available options below:</em>
<em>a) 2.97
</em>
<em>b) 3.10
</em>
<em>c) 3.19
</em>
<em>d) 3.22
</em>
<em>e) 3.32</em>
The correct option is b.
Explanation:
firstly firstly we are to calculate the number of moles of ammonia and using the mole concept of two moles of ammonia gives one mole of ammonium sulphate we can calculate the number of moles of ammonium sulphate and mass
from n=m/mr
Answer:
Explanation:There are no results for balance the following equation:ni(cio), →nici,+0when the equation is properly balanced, the correctcoefficients are, respectively
Answer:
a) 1.71 × 10⁻³ M
b) 8.00 × 10⁻⁵ M
Explanation:
In order to calculate the solubility (S) of Pb(SCN)₂ we will use an ICE chart. We identify 3 stages (Initial, Change, Equilibrium) and complete each row with the concentration or change in the concentration.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product (Ksp) is:
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (2S)² = 4S³
S = 1.71 × 10⁻³ M
<em>b) Calculate the molar solubility of lead thiocyanate in 0.500 M KSCN.</em>
KSCN is a strong electrolyte that dissociates to give 0.500 M K⁺ and 0.500M SCN⁻.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0.500
C +S +2S
E S 0.500 + 2S
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (0.500 + 2S)²
In the term (0.500 + 2S)², 2S is negligible.
Ksp = 2.00 × 10⁻⁵ = S . (0.500)²
S = 8.00 × 10⁻⁵ M
