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seraphim [82]
3 years ago
5

A 25.0 ml sample of an unknown monoprotic acid was titrated with 0.12 m naoh. the student added 31.6 ml of naoh and went past th

e equivalence point. which procedure could be performed next to more accurately determine the concentration of the unknown acid?
Chemistry
1 answer:
Inessa05 [86]3 years ago
7 0
The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants. 
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
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Of the five salts listed below, which has the highest concentration of its cation in water? assume that all salt solutions are s
Zigmanuir [339]
First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:

Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively. 

For the solutions ahead, let x be the concentration of the cation.

A.  The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M

B. The formula is Co(OH)₂. 
 1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M

C. The formula is CoS. 
 5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M

D. The formula is Cr(OH)₃. 
 1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M

E. The formula is Ag₂S. 
 6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M

<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>

5 0
3 years ago
How many moles of nitrogen are contained in 18.65 L at STP?
stira [4]
1 mole of any gas under STP  ----- 22.4 L

18.65 L*1 mol/22.4 L ≈ 0.8326 mol N2
8 0
3 years ago
Read 2 more answers
Hydrazine, N2H4, is a corrosive liquid used in rocket and jet fuels. Ammonia, NH3, is a gas that dissolves in water to form a so
Lyrx [107]

OK in the case of hydrazine  14 grams of nitrogen  combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.

Ratio 2:3
5 0
3 years ago
How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3}  }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}

3 0
4 years ago
Mass divided by volume *
lorasvet [3.4K]
D — density

hope this helps !
5 0
3 years ago
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