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stepladder [879]
3 years ago
11

Write the formulas of all species in solution for the following ionic compounds by writing their dissolving equations:

Chemistry
1 answer:
Naya [18.7K]3 years ago
7 0

Answer:

1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

Explanation:

Let's consider the dissolving equations for the following compounds.

1. Rubidium hydroxide

RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)

2. Sodium carbonate

Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)

3. Ammonium selenite

(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)

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Is it possible for the equivalence point of a titration to not be at pH 7? Explain your answer.
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<span>The reason it will be 7 for some titrations is that when you  titrates a strong acid with a strong base for example  HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
 That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7 
second reason is 
When you titrates a weak acid with strong base at equivalence point 
only a water solution of the conjugate base exists 

CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O 
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so 
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>
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3 years ago
Which alkaline earth metal is a component of gypsum​
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6 0
3 years ago
what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

#SPJ4

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