Can someone show how to answer these 3 ph questions? (dont just give answers plz
ID: 852639 • Letter: C
Question
Can someone show how to answer these 3 ph questions? (dont just give answers plzzz)
(1)
What is the pH of a solution of 0.81 M acid and 0.35 M of its conjugate base if the acid dissociation ionization constant, Ka is 5.45 x 10-8?
(2)
A 20.0 ml sample of 0.100 M CH3COOH is titrated with 0.100 M NaOH. Calculate the pH of the solution at points where 15.00 ml of NaOH have been added. (Ka for CH3 COOH = 1.8x10-5)
(3)
Ka for hypochlorous acid, HClO, is 3.0x10-8. Calculate the pH after addition of 15.0 , 20.0 , 40.0 , and 50.0 ml of 0.100 M NaOH to 40.0 ml of 0.100 M HClO. Identify the pH at equilibrium and half equilibrium.
Explanation / Answer
2)
number of moles of CH3COOH = 0.100M *( 1L*20.0mL /1000mL)=0.002moles
number of moles of NaOH=0.100M *(1L*15mL/1000mL)=0.0015mole
the reaction between CH3COOH and NaOH is
CH3COOH(aq) + NaOH (aq)--- > CH3COONa (aq)+ H2O(l)
I: 0.002 0.0015 0.0000
C: - 0.0015 -0.0015 +0.0015
E:(0.002 - 0.0015) 0.0000 +0.0015
total volume is 35mL is same for both CH3COOH and CH3COONa, we can consider moles as concentration.
Ka=[CH3COONa][H^+]/[CH3COOH]
1.8*10^-5 =(1.5*10^-3)*[H^+]/0.5*10^-3
[H^+] = 0.6*10^-5
pH=-log[H^+]
pH=- log( 0.6*10^-5) = 5.2218