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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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1 year ago

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$

So the total mole of chloride ions:

$N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-$

And the total volume by adding the volume of each solution in L:

$V=0.500L+0.425L=0.925L$

Finally, the molarity turns out:

$M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M$

Best regards.

5 0

2 years ago

What is the stoichiometric ratio between BaCl2 and NaCl

bixtya [17]

BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.

7 0

2 years ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Explanation:

For 1: At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

For 2:

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$

So, $1.7\times 10^{-4}mol$ of methyl amine will produce = $\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+$

To calculate the $pK_b$ of base, we use the equation:

$pK_b=-\log(K_b)$

where,

$K_b$ = base dissociation constant = $3.9\times 10^{-10}$

Putting values in above equation, we get:

$pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41$

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})$

We are given:

$pK_b=9.41$

$[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

$[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

Putting values in above equation, we get:

$pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-9.41=4.59$

Hence, the pH of the solution is 4.59

4 0

2 years ago

A sample of n2 gas occupies 2.40 l at 20°c. if the gas is in a container that can contract or expand at constant pressure, at wh

Bas_tet [7]

From Charle's law the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant pressure.
Therefore';
V1/T1=V2/T2
Where; V1 = 2.40 l, T1 = 273 +20= 293 K, V2 = 4.80, and T2= ?
2.4/293= 4.8/T2
T2= (4.8×293)/2.4
= 586 K or 313° C

4 0

2 years ago

Read 2 more answers