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2 years ago

Which is the stronger acid in each of the following pairs? Explain your reasoning.

Chemistry

1 answer:

Makovka662 [10]2 years ago

6 0

Answer:p-hydroxybenzaldehyde is stronger acid to phenol

para-cyanophenol is stronger acid to meta-cyanophenol

o-fluorophenol is stronger acid to p-fluorophenol.

Explanation:

The PKa tool relative to Ph are used to contrast the pairs.

The pKa of phenol is 10. The pKa of p-hydroxybenzaldehyde is 9.24

The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95.

The pKa value of o-fluorophenol is 8.7, while that of the p-fluorophenol is 9.9. It's obvious that the inductive effect is more dominant at ortho-position, which results in a more acidic nature

The pKa is the pH value at which a chemical species will accept or donate a proton. The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.

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A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an

Assoli18 [71]

a.
Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side, Right hand side,
H atoms = 2   H atoms = 2
Cl atoms = 1   Cl atoms = 1
Na atoms = 1   Na atoms = 1
O atoms = 1   O atoms = 1

Hence, the reaction equation is already balanced.

b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.

Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M

6 0

2 years ago

Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As

Crank

Answer:

1.8 × 10⁻¹⁶ mol

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

1.2 + 3s 2s

$K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\$

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0

2 years ago

When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C

UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

$[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M$

Equilibrium constant of the reaction :

$K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}$

K = 4

$CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)$

Concentration at eq'm:

0.24 M 0.24 M 0.48 M 0.48 M

After addition of 0.34 moles per liter of $CO_2$ and $H_2$ are added.

(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

$K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}$

$4=\frac{(0.48+x)^2}{(0.24+0.34)^2}$

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0

2 years ago

Which of these correctly defines the pOH of a solution? the log of the hydronium ion concentration the negative log of the hydro

Likurg_2 [28]

Answer : The correct option is, the negative log of the hydroxide ion concentration.

Explanation :

pOH : It is defined as the negative logarithm of hydroxide ion concentration. It is a measure of the alkalinity of the solution.

Formula used :

$pOH=-log[OH^-]$

$[OH^-]$ is the concentration of $OH^-$ ions.

When pOH is less than 7, the solution is alkaline.

When pOH is more than 7, the solution is acidic.

When pOH is equal to 7, the solution is neutral.

4 0

2 years ago

Read 2 more answers

Determine the number of moles in 4.21 x 10^23 molecules of CaCl2

Paha777 [63]

<h3>Answer:</h3>

0.699 mole CaCl₂

<h3>Explanation:</h3>

To get the number of moles we use the Avogadro's number.

Avogadro's number is 6.022 x 10^23.

But, 1 mole of a compound contains 6.022 x 10^23 molecules

In this case;

we are given 4.21 × 10^23 molecules of CaCl₂

Therefore, to get the number of moles

Moles = Number of molecules ÷ Avogadro's constant

= 4.21 × 10^23 molecules ÷ 6.022 x 10^23 molecules/mole

= 0.699 mole CaCl₂

Hence, the number of moles is 0.699 mole of CaCl₂

7 0

2 years ago