Answer: hydroxide ions
Explanation:
According to the Arrhenius concept, an acid is a substance that ionizes in the water to give hydronium ion or hydrogen ion and a bases is a substance that ionizes in the water to give hydroxide ion .
According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
As KOH can give hydroxide ions on dissociation , it is considered as arrhenius base.
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
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1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
To determine what gas is this, we use Graham's Law of Effusion where it relates the rates of effusion of gases and their molar masses. We do as follows:
r1/r2 = √(M2 / M1)
Let 1 be the the unkown gas and 2 the H2 gas.
r1/r2 = 0.225
M2 = 2.02 g/mol
0.225 = √(2.02 / M1)
M1 = 39.90 g/mol
From the periodic table of elements, most likely, the gas is argon.
