Question

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*Be careful with your significant digits*

For a 0.36 M solution of HNO₃, calculate: [H⁺], pH, pOH, and [OH^-]

Determine all pH and pOH values to two decimal places.

1) To enter an answer in scientific notation: 1.2×10⁻⁴ = 1.2E-4

[H⁺]= 0.36 M
pH= 0.44
pOH= 13.56
[OH^-]= 2.7×10^-14M

You are correct. Your receipt no. is 164-4624

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2) Now make the same calculations for a 1.0×10⁻¹³ M solution of HI.

[H⁺]=  
pH=
pOH=
[OH^-]=  

Incorrect.

Answer 1

pH = -log [H+]

pOH = -log [OH-]

pH + pOH = 14

Kw = [H+] [OH-]

Where Kw= 10E-14

HI - 1.0x1o-13 M [HI] = 1,0x10-13 M pH = -log[he E-log (160 x 10") = 13 log 1o - log 1 z 13 tous S leglo= leg 1 to PH e pol=14 o poh=14 - 13=1n [OH-] = Kw = 1014 H] X2 10-13 10 2M