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3 years ago

What is the ph of a 0.027 M KOH solution?

1 answer:

8 0

First you calculate the concentration of [OH⁻] in <span>solution :

POH = - log [ OH</span>⁻]

POH = - log [ 0.027 ]

POH = 1.56

PH + POH = 14

PH + 1.56 = 14

PH = 14 - 1.56

PH = 12.44

hope this helps!

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Gemiola [76]

<h3>Answer:</h3>

28 mol CaF

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 28.2298 \ moles \ CaF$

<u>Step 4: Check</u>

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

28.2298 mol CaF ≈ 28 mol CaF

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Read 2 more answers

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