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dalvyx [7]
3 years ago
13

Calculate the pH at of a 0.10 Msolution of anilinium chloride . Note that aniline is a weak base with a of . Round your answer t

o decimal place. Clears your work. Undoes your last action. Provides information about entering answers.
Chemistry
1 answer:
notsponge [240]3 years ago
8 0

The question is incomplete, here is the complete question:

Calculate the pH at of a 0.10 M solution of anilinium chloride (C_6H_5NH_3Cl) . Note that aniline (C6H5NH2) is a weak base with a pK_b of 4.87. Round your answer to 1 decimal place.

<u>Answer:</u> The pH of the solution is 5.1

<u>Explanation:</u>

Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).

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

pH=7-\frac{1}{2}[pK_b+\log C]

where,

pK_b = negative logarithm of weak base which is aniline = 4.87

C = concentration of the salt = 0.10 M

Putting values in above equation, we get:

pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1

Hence, the pH of the solution is 5.1

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3 0
3 years ago
Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3
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<u>Answer:</u> The volume of barium chlorate is 195.65 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL

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6 0
3 years ago
According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra
Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 =  0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

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