Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
Answer:
HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)
HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)
HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)
HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)
Explanation:
Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.
- When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).
- When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).
Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.
- When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).
- When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).
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Ba²⁺ + 2Cl⁻ + 2H⁺ + SO₄²⁻ = BaSO₄ (precipitate) + 2H⁺ + 2Cl⁻
Ba²⁺ + SO₄²⁻ = BaSO₄