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alisha [4.7K]
3 years ago
7

What is the ph of a solution of 0.650 m k2hpo4, potassium hydrogen phosphate?

Chemistry
1 answer:
Pie3 years ago
3 0

phosphate?

At 25 ∘C phosphoric acid, H3PO4, has the following equilibrium constants:

Ka1= 7.5×10^-3

Ka2= 36.2×10^−8

Ka3 = 4.2×10^−13

This is a question about the dissociation of weak polyprotic acids.

When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2

pKa2 = -log(Ka2) ; pKa3 = -log(Ka3)

So, for this problem, REGARDLESS OF THE SUPPLIED CONCENTRATION , the answer is:

pH = (7.2076+12.3767) / 2

pH = 9.79

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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.2 \cdot 10^{24} \ molecules \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{6.022 \cdot 10^{23} \ molecules \ C_6H_{12}O_6})
  2. Divide:                                                                                                                      \displaystyle 1.99269 \ mol \ C_6H_{12}O_6

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

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