Q. 484.5( 8 Votes )

Assuming complete

Answer :

To calculate the pH, we apply the formula:

For acidic solution, pH = -log[H3O+]


For basic solution, pOH = - log[OH-]


Also, we use the given formula to calculate pH and pOH


pH + pOH = 14


A)0.003M HCl


Given:


[HCl] = 0.003 M


Ionization of HCl


HCl + H2O H3O+ + Cl-


Since HCl is completely ionized(given)


[HCl] = [H3O+]


[H3O+] = 0.003 M


By applying the formula,


pH = -log[H3O+]


As [H3O+] = 0.003 M


pH = -log[0.003]


pH = 2.52


Thus, the pH of the solution is 2.52


B) 0.005 M NaOH


Given:


[NaOH] = 0.005M


Ionization of NaOH,


NaOH Na+ + OH-


Since NaOH is completely ionized (given)


[NaOH] = [OH-]


[OH-] = 0.005 M


By applying the formula for basic solution:


pOH = -log[OH-]


As [OH-] = 0.005 M


pOH = -log [0.005]


pOH = 2.30


As pOH + pH =14


pH = 14 - pOH


pH = 14 – 2.30


pH = 11.70


Thus, the pH of the solution is 11.70


c)0.002 M HBr


Given:


[HBr] = 0.002 M


Ionization of HBr


H2O + HBr H3O+ + Br-


Since HBr is completely ionized(given)


[HBr] = [H3O+]


[H3O+] = 0.002 M


By applying the formula (for acidic solution)


pH = -log[H3O+]


As [H3O+] = 0.002 M


pH = -log[0.002]


pH = 2.69


Thus, the pH of the solution is 2.69


d)0.002 M KOH


Given:


[KOH] = 0.002M


Ionization of KOH


KOH + K+ + OH-


Since KOH is completely ionized (given)


[KOH] = [OH-]


[OH-] = 0.002 M


By applying the formula for basic solution,


pOH = -log[OH-]


As [OH-] = 0.002 M


pOH = -log[0.002]


pOH = 2.69


As pOH + pH =14


pH = 14 - pOH


pH = 14 – 2.69


pH = 11.31


Thus, the pH of the solution is 11.31


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