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Cerrena [4.2K]
2 years ago
14

What is the ph of a solution made by mixing 30.00 ml of 0.10 m acetic acid with 50.00 ml of 0.100 m koh? assume that the volumes

of the solutions are additive. k a = 1.8 × 10-5 for ch3co2h.
SAT
1 answer:
morpeh [17]2 years ago
5 0

Given 30.00 ml of 0.10 m acetic acid and 50.00 ml of 0.100 m koh, the pH of the solution is 12.40

<h3>What is the pH of a substance?</h3>

This is the measure of the acidity or the basicity of an aqueous or liquidsolution.

The details that are contained in this question are:

Mole of CH3COOH = 0.1 M

Volume of CH3COOH  = 30 mL

Mole of KOH = 0.1 M

Volume of KOH = 50 mL

mol of (CH3COOH) = 0.1 Mole * 30 mL = 3 mol

mol of (KOH) = 0.1 Mole * 50 mL = 5 mol

From the solution above,

mol of CH3COOH = 3 mol

mol of KOH = 5 mol

3 mol of both substances is going to react

excess KOH remaining = 5 - 3 =  2 mmol

Volume of both Solutions

= 30 + 50 = 80 mL

OH⁻ = 2/80

= 0.025 M

use:

pOH = -log [OH-]

= -log (0.025)

pOH = 1.602

PH = 14 - pOH

= 14 - 1.60

= 12.40

The calculated pH is therefore 12.40

Read more on pH here: brainly.com/question/22390063

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If a person takes 30 ml of vinegar a day and the molarity of the vinegar is 0. 84 m then how many grams of acetic acid (hc2h3o2
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The mass of the acetic acid is 1.512 g.

<h3 /><h3>What is Molarity?</h3>

molarity is the concentration of a substance in mol per dm³

From the question,

The number of moles contained in the 30 mL vinegar is

  • n = (0.84×30)/1000
  • n = 0.0252 mol.

the molar mass of vinegar is calculated below.

  • m.m of (HC₂H₃O₂) = [1+(12×2)+(1×3)+(16×2)]
  • m.m = (1+24+3+32]
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To calculate the mass of acetic acid, we use the formula below.

  • R.m = n×m.m............. Equation 1

Where

  • R.m = mass of acetic acid.

Substitute these values into equation 1

  • R.m = (0.0252×60)
  • R.m = 1.512 g.

Hence, The mass of the acetic acid is 1.512 g.

Learn more about molarity here: brainly.com/question/20383725

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