Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
23 would be the answer actually idk I just need to answer some answers
Answer:
10.945 x 10^-4
Explanation:
Balanced equation:
Mn(OH)2 + 2 HCl --> MnCl2 + H2O
it takes 2 moles HCL for each mole Mn(OH)2
Next find the molarity of the Mn(OH)2 solution
= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)
= 4.86 x 10^-3 mole
this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L
thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M
Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4
<u>Answer:</u> The correct answer is 0.745 km
<u>Explanation:</u>
We are given:
A numerical value of 815 yards
To convert this into kilometers, we use the conversion factor:
1 km = 1093.6 yards
Converting the given value into kilometers, we get:
Hence, the correct answer is 0.745 km
Answer: 1.32
Explanation:
First, we must obtain the molar mass of HBr. After that, we try to obtain the concentration of the hydrobromic acid from the formula n=CV since the volume of solution and mass of acid was provided. Recall that n=m/M. If the concentration of acid is thus obtained, we make use of the fact that the concentration of H+ in the acid is equal to the molar concentration of HBr to obtain the pH. The pH is the negative logarithm of the concentration we obtained in the initial step.
