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Sati [7]
3 years ago
15

Identify two acids and two bases that you use or come into contact with in an average week. Identify uses for each substance.

Chemistry
2 answers:
skelet666 [1.2K]3 years ago
4 0

Explanation:

Two acids we come into contact with in an average week

  1. Vinegar is an 10% solution of acetic acid (CH_3COOH)in water. Used in salad dressing and while cooking food. It has a sour taste.
  2. Citric acid present in fruits and vegetables like : lemons, orange, tomatoes etc. It is a weak organic acid with sour taste.

Two bases we come into contact with in an average week.  

  1. Baking soda (NaHCO_3) is used in baking food like: cakes, cookies, breads. Baking soda is one of the ingredient while baking breads and cakes.
  2. Caustic soda (NaOH) is used for preparation of detergents, papers , soaps etc. We use soaps and detergents for washing.
DaniilM [7]3 years ago
4 0
1 common acid that is used on a daily basis is Hydrochloric Acid in your stomach. It actives the enzyme pepsin into pepsinogen to degenerate proteins. Another acid is vinegar which is used in cooking and cleaning. 

A base I use on a daily basis is Tums as it has sodium bicarbonate, a common base used to neutralize stomach acids. Another one is ammonia which is used for hair and cleaning products. 
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Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueou
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Answer:

[K⁺] = 0.107 M

[OH⁻] = 1.13 ×  10⁻⁹ M

Explanation:

600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻

⇒ 0.45 × 600 × 10⁻³

= 0.27 moles of Cu²⁺ and (NO₃)²⁻

450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻

⇒ 0.25 × 450 × 10⁻³

= 0.1125 moles of K⁺ and OH⁻

Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺  (because 1 Cu²⁺  needs 2 OH⁻)

Therefore , moles of remaining Cu²⁺  = 0.27 - 0.05625

=0.21375 moles which is equal to :

⇒ 0.21375/(( 600+450))× 10⁻³

= 0.21375/1050 × 10⁻³

= 0.20357 M

Given that :

(Ksp for Cu(OH)2 is 2.6 ×  10⁻¹⁹)

We know that , Ksp = [Cu²⁺][OH⁻]²

2.6 ×  10⁻¹⁹ = 0.20357 × [OH⁻]²

[OH⁻]² = 2.6 ×  10⁻¹⁹/0.20357

[OH⁻] = 1.13 ×  10⁻⁹ M

[K⁺] = moles of K⁺ /total volume

[K⁺] = 0.1125 / 1050 × 10⁻³

[K⁺] = 0.107 M

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