Answer:
Explanation:
Given parameters:
pH = 3.50
Unknown:
concentration of [H₃0⁺] = ?
concentration of [OH⁻] = ?
Solution:
In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.
pH = -log₁₀[H₃O⁺]
[H₃O⁺] = inverse log₁₀ (-pH) =
= 
[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³
For the [OH⁻]:
we use : pOH = -log₁₀ [OH⁻]
Recall: pOH + pH = 14
pOH = 14 - pH = 14 - 3.5 = 10.5
Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]
[OH⁻] = 
[OH⁻] =
= 3.2 x 10⁻¹¹moldm⁻³
The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.
Answer:
3M
Explanation:
moles ÷ liters = molarity
4.8 ÷ 1.6 = 3M
Answer:
the concentration of the reactants
the temperature in heating
Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially =
moles = 0.00600 moles
Neutralization reaction (back titration): 
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration =
moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.