Respuesta :
Answer:
[tex][H^+]=0.000285[/tex]
[tex]pH=3.55[/tex]
Explanation:
In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:
[tex]HN_3~<->~H^+~+~N_3^-[/tex]
Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:
[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]
For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:
[tex]Ka=\frac{X*X}{[HN_3]}[/tex]
Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:
[tex]Ka=\frac{X*X}{0.004-X}[/tex]
Finally, we can put the ka value and solve for "X":
[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]
[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]
[tex]X= 0.000285[/tex]
So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:
[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]
I hope it helps!
The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and 3.527759
pH based problem:
What information do we have?
Hydrazoic acid solution = 0.0040 M
Ka of hydrazoic acid = 2.20 × 10⁻⁵
We know that weak acids
[H+] = √( Ka × C)
[H+] = √( 2.2 × 10⁻⁵ × 0.0040)
[H+] = 0.000296648
So,
pH = -log [H+]
pH = -log [0.000296648]
Using log calculator
pH = 3.527759
Find more information about 'pH'.
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