The molarity of the HCl is 1 M when 12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution.
Explanation:
Data given:
molarity of the base NaOH, Mbase =0. 5 M
volume of the base NaOH, Vbase = 12 ml
volume of the acid, Vacid = 6 ml
molarity of the acid, Macid = ?
The titration formula for acid and base is given as:
Mbase Vbase = Macid Vacid
Macid =
Macid = 1 M
we can see that 1 M solution of HCl was used to neutralize the basic solution of NaOH. The volume of NaOH is 12 ml and volume of HCl used is 6ml.
Answer:
Acids have a high concentration of [H⁺] ions
Explanation:
In order to know about the acids, a solution is acidic when pH < 7
As pH is lower than 7, the pOH > 7
When pH = 7, solution is neutral
When pH is greater than 7, solution is basic
pH = - log [H⁺]
pOH = - log [OH⁻]
Imagine a solution of pure HCl 0.2 M
HCl → H⁺ + OH⁻
[H⁺] = 0.2 M → pH = - log 0.2 → 0.69
pH + pOH = 14
pOH = 14 - 0.69 = 13.31
10^-pOH = [OH⁻] → 10⁻¹³°³¹ = 4.89ₓ10⁻¹⁴
In conclussion [H⁺] > [OH⁻]
PH is the test of acidity or basicity of a solution. it follows the formula:
pH = pKa + log [salt] / [acid] where NaF is the salt and HF is the acid in this case.
By literature, Ka of HF is 3.5*10^-4
<span>pKa= -log(Ka)=</span><span> 3.46 </span>
<span>pH = pKa + log [NaF / [HF] </span>
4.05 = 3.46 + log [NaF / [HF]
log [NaF / [HF]<span> = 0.59
</span>
[NaF / [HF] = 3.89
Answer:
Agree
Explanation:
The way it makes electicity is that it rubs together fast enother it makes the particles move.
Sorry I can't do better
Answer:
the answer is Fe + 02 - Fe203
Explanation:
