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arsen [322]
3 years ago
13

Consider a solution prepared by mixing the following:50.0 mL of 0.100 M Na3PO4100.0 mL of 0.0500 M KOH200.0 mL of 0.0750 M HCl50

.0 mL of 0.150 M NaCNDetermine the volume of 0.100 M HNO3 that must be added to this mixture to achieve a nal pH value of 7.21.
Chemistry
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

you must add 50 mL

Explanation:

Hi

KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.

NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.

HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.

The acid reacts with the bases leaving 2.5 mmol unreacted.

Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.

The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12

5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21

Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.

To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.

successes with your homework

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You titrate 25.00 mL of the lemon-lime Kool-Aid (with KI, HCl, and starch) with 0.001000 M KIO3(aq) solution. The titration requ
Svetradugi [14.3K]

Answer:

1.0190 x 10⁻⁵ mol

Explanation:

We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).

Molarity = mol/V

V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L

⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L =  1.0190 x 10⁻⁵ mol KIO₃

# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol

8 0
3 years ago
A certain weak acid, ha, has a ka value of 3.6×10−7. part a calculate the percent ionization of ha in a 0.10 m solution. express
slega [8]
Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
8 0
3 years ago
Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M
liberstina [14]

Answer:

= \mathbf{C_3H_6O}

Explanation:

From the given information, since the molecular mass of the ion M+ is not given;

Let's assume M+ = 58.0423

So, by applying the 13th rule;

we will need to divide the mass by 13, after dividing it;

The quotient n = no. of carbon; &

The addition of the quotient (n) with the remainder r =  no. of hydrogen.

So;

\dfrac{58}{13}= 4 \ remainder \ 6

So;

C_nH_{n+r} = C_4H_{4+6}

= C_4H_{10}

From the given information; we have oxygen present, so since the mass of oxygen = 16, we put oxygen in the molecular formula by removing CH_4. Also, since the mass is an even number then Nitrogen is 0.

So, we have:

= \mathbf{C_3H_6O}

6 0
2 years ago
Baking soda NaHCO3, is made from soda ash, a common name for sodium carbonate. The soda ash is obtained in two ways. It can be m
Vaselesa [24]

Answer:

Explanation:

From the given information:

mass of silver chloride AgCl = 11.89 g

molar mass of AgCl = 143.37 g/mol

We know that:

number of moles = mass/molar mass

∴

number of moles of AgCl = 11.89 g/ 143.37 g/mol

number of moles of AgCl = 0.0829 mol

The chemical equation for the mineral called trona is:

\mathsf{Na_2CO_3.NaHCO_3.2H_2O}

when being reacted with hydrochloric acid, we have:

\mathsf{Na_2CO_3.NaHCO_3.2H_2O + 3HCl \to 3NaCl + 2CO_2 +4H_2O}

One mole of NaCl formed from one mole of trona sample = 0.0829 moles of AgCl

i.e. 0.0829 moles of NaCl can be formed from AgCl

mass of trona sample = number of moles × molar mass

mass of trona sample = 0.0829 × 226

mass of trona sample = 18.735 g

The mass in the percentage of NaHCO₃ = mass of NaHCO₃/ mass of trona

The mass in the percentage of NaHCO₃ = 6.93/18.735

The mass in the percentage of NaHCO₃ = 0.36989

The mass in the percentage of NaHCO₃ = 36.99%

Nonetheless, a 6.78 g samples manufactured from sodium carbonate in pure 100%

∴

6.78 g sample manufactured from Na₂CO₃ is purer.

8 0
3 years ago
2) Convert 2.65*10^ ^ 25 atoms of Chlorine to moles of CI
Irina18 [472]

Answer:

22 mol

Explanation:

Given data:

Number of atoms of Cl = 2.65×10²⁵ atom

Number of moles of Cl = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

2.65×10²⁵ atom × 1 mol / 6.022 × 10²³ atoms

0.44×10² mol

22 mol

7 0
3 years ago
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