The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.

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04-05-2019
Chemistry

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The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.

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superman1987 superman1987 · Answer 1 · 2019-06-05T15:54:14+02:00

Answer:

12.6.

Explanation:

We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-03-16T09:09:03+01:00

12.6 is the pH of a solution prepared by mixing 45.0 ml of 0.183 m KOH and 35.0 ml of 0.145 m HCl.

How we calculate pH?

pH of the solution can be calculated as pH = 14 - pOH.

First of all we have to calculate the concentration of OH⁻ ions as follow:

Total volume of the solution = 45ml + 35ml = 80ml

Concentration of OH⁻ in terms of mole is calculated as:

[OH⁻]=(no. of millimoles of KOH - no. of millimoles of HCl)/Total volume

Given that, concentration of KOH = 0.183m

Volume of KOH = 45ml

Concentration of HCl = 0.145m

Volume of HCl = 35ml

Millimoles can be calculated as:

millimoles = concentration × volume

No. of millimoles of KOH = 0.183 M × 45.0 mL = 8.235 mmol

No. of millimoles of HCl = 0.145 M ×35.0 mL = 5.075 mmol

Now, [OH⁻] = 8.235 mmol - 5.075 mmol / 80.0 mL = 0.395 M

pOH = -log[OH⁻]

pOH = -log(0.395)

pOH = 1.4

Therefore, pH = 14 - 1.4 = 12.6

Hence, 12.6 is the pH of the solution.

To know more about pH, visit the below link:

https://brainly.com/question/26767292