Question

What is the conjugate acid of HPO32− ?

Part D The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 7.00 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH, note the corresponding OH− ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L)" above the graphic. Select on the Logarithmic scale below the graphic.

Find the pOH of the solution.

Answer 1

Answer:

Given,

To determine the pOH

we know,

pH is defined as the negative logaritm of H+ion concentration

i.e.,

pH = -log [H+]

7 = -log [H+]

[H+] = 10^-7

[H+] = 1*10^-7

now

consider,

[H+] [OH-] = 10^-14

so

1 x 10^-7 x [OH-] = 10^-14

[OH-] = 10^-14 / 10^-7

[OH-] = 1*10^-7

so

Concentration of OH- i.e.. [OH-] = 1*10^-7

pOH = -log [OH-]

pOH = - log (1*10^-7)

pOH = 7