Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.
Answer:
Explanation:
Unit 10 - Acid/Base ... (a) Mg(OH. 2. ) (b) Mg(OH). 2. (c) Mg. 2. OH. (d) MgOH. 2. Standard: ... balanced equation for these neutralization reactions: 3. HCl + NaOH → ... H2CO3 + Ca(OH)2 → ... C5.7B Predict products of an acid-base neutralization. 8. 2 NH4OH + H2S ...An Arrhenius base is a compound that increases the OH − ion concentration in ... and a base is called a neutralization reaction and can be represented as follows: ... chemical equation for the neutralization reaction between HCl and Mg(OH) 2. ... acid, an Arrhenius base, or neither. a) NaOH. b) C 2H 5OH. c) H 3PO 4. 6
Answer:
0.40 g/cm3
Explanation:
density = mass / volume.
mass = 65.2 grams
volume = 10*1.1*15=165 cm3
so density = 65.2/165=0.40 g/cm3
The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.
