[H+][OH-]=10-¹⁴
We are using this formula because we need to find the H+
substitute the value given for hydronium ion for OH-
[H+][4.19×10⁵]=10-¹⁴
[H+]=10-¹⁴÷4.19×10⁵
[H+]=2.387×10-¹⁹
Then the pH of the solution will be
pH= –log¹⁰ [H+]
pH = –log¹⁰ [2.387×10-¹⁹]
pH= –log¹⁰2.387+19log¹⁰
= –0.378+19
pH =18.622
Answer:
pH = 9: Basic; pH = 2.3: acidic; pH = 11: basic; pH = 5: Acidic; [H⁺] = 0.0056M: Acidic; [H⁺] = 3.45E-9M: Basic
Explanation:
A solution is defined as acidic when pH < 7 and as basic when pH > 7.
Also, pH = -log[H⁺].
Thus:
pH = 9: >7 → Basic
pH = 2.3: <7 → Acidic
pH = 11: >7 → Basic
pH = 5: <7 → Acidic
[H⁺] = 0.0056M, pH = -log0.0056M = 2.25: <7: Acidic
[H⁺] = 3.45E-9M, pH = 8.46: > 7: Basic
The concentration of this acid in moles per litre : 17.41
<h3>Further explanation</h3>
Given
99.5% acetic acid
density=ρ=1.05 g/ml
Required
the concentration (mol/L)
Solution
99.5% dan density 1.05 g/ml (MW acetic acid = 60 g/mol)
![\tt 99.5\%\times 1.05\dfrac{g}{ml}\times \dfrac{1000~ml}{L}=1044.75~g/L\\\\\dfrac{1044.75~g/L}{60~g/mol}=17.41~mol/L](https://tex.z-dn.net/?f=%5Ctt%2099.5%5C%25%5Ctimes%201.05%5Cdfrac%7Bg%7D%7Bml%7D%5Ctimes%20%5Cdfrac%7B1000~ml%7D%7BL%7D%3D1044.75~g%2FL%5C%5C%5C%5C%5Cdfrac%7B1044.75~g%2FL%7D%7B60~g%2Fmol%7D%3D17.41~mol%2FL)
F2 equals 10 N and F4 equals 2 N
There are four naturally occurring isotopes of iron, Fe-54, Fe-56, Fe-57, and Fe-58.