Respuesta :
You can use the fallowing equations:
pH=-log[H⁺]
pH=14-pOH
pOH=-log[OH⁻]
[H⁺]=K(w)/[OH⁻] (K(w)=1.01×10⁻¹⁴)
You can use any of those equation in what ever order to find the [H⁺] and then pH. If pH>7, the solution is basic. If the pH<7, the solution is acidic.
I will show you how I would find it but there is more than one way to do it.
[H⁺]=(1×10⁻¹⁴)/(10⁻¹¹)
[H⁺]=0.001M
pH=-log(0.0010)
pH=3.00
Therefore the solution is acidic.
I hope this helps. Let me know in the comments if anything is unclear.
pH=-log[H⁺]
pH=14-pOH
pOH=-log[OH⁻]
[H⁺]=K(w)/[OH⁻] (K(w)=1.01×10⁻¹⁴)
You can use any of those equation in what ever order to find the [H⁺] and then pH. If pH>7, the solution is basic. If the pH<7, the solution is acidic.
I will show you how I would find it but there is more than one way to do it.
[H⁺]=(1×10⁻¹⁴)/(10⁻¹¹)
[H⁺]=0.001M
pH=-log(0.0010)
pH=3.00
Therefore the solution is acidic.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
If we have [tex][H+][OH-] = Kw = 1.0 x 10^-^1^4[/tex]
Then [tex][H+]= Kw/ [OH-]= 1.0x 10^-^1^4/ 1 x 10^-^1^1 =1 x 10^-^3 mol/L[/tex]
[tex]pH = - log [H+] = - log 1 x 10^-^3 = 3[/tex] [tex]< 7[/tex]
This is a Acidic Solution
hope this helps
give brainliest if you please