07-13: The diprotic acid Hal. has Kar-1.4 x 10 and Kaz -1.2 x 10-10 97-Calculate the molarity of H" of 0.080 M HL solution 13-10¹ B) 4.1-10 C) 1.3×10% D) 4.1 × 107 Q8-Calculate the molarity of Lof 0.080 M H₂L solution A) 1.4-10 B) 1.2×10 C) 1.4×10 D) 1.2 × 10-10 09- Find the pH of a solution prepared by dissolving 0.85 g of NaHL (188.1 g/mol) and 1.05 g of NaL (194.0 g/mol) in 250.0 mL of water. A) 10.84 B) 9.00 C) 7.84 D) 10.0
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
pleasssse solve parts 7, 8 and 9
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### Acid-Base Chemistry Problems
The diprotic acid H₂L has \(K_{a1} = 1.4 \times 10^{-6}\) and \(K_{a2} = 1.2 \times 10^{-10}\).
**Q7-13**:
**Q7.** Calculate the molarity of \(H^+\) of 0.080 M HL⁻ solution
A) \(1.3 \times 10^{-6}\)
B) \(4.1 \times 10^{-6}\)
C) \(1.3 \times 10^{-8}\)
D) \(4.1 \times 10^{-7}\)
**Q8.** Calculate the molarity of \(L^{2-}\) of 0.080 M H₂L solution
A) \(1.4 \times 10^{-6}\)
B) \(1.2 \times 10^{-6}\)
C) \(1.4 \times 10^{-5}\)
D) \(1.2 \times 10^{-10}\)
**Q9.** Find the pH of a solution prepared by dissolving 0.85 g of NaHL (molar mass = 188.1 g/mol) and 1.05 g of Na₂L (molar mass = 194.0 g/mol) in 250.0 mL of water.
A) 10.84
B) 9.00
C) 7.84
D) 10.0
**Q10.** The fraction of \(L^{2-}\) (\(\alpha\) of \(L^{2-}\)) at pH = 12.00 is
A) 0.923
B) 0.813
C) 0.992
D) 0.089
**Q11.** At pH = 5.85
A) [HL⁻] = [H₂L]
B) [L²⁻] > [HL⁻]
C) [HL⁻] > [H₂L]
D) [L²⁻] = [HL⁻]
**Q12.** At pH = 7.89
A) [HL⁻] = [H₂L]
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