Solution check for pH of Strong Acid and Strong Base

  • Thread starter pavadrin
  • Start date
  • #1
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Homework Statement


Calculate the pH after 24 mL of 1.00 M NaOH is added from the burette to a beaker containing 25.0 mL of 1.00 M HCl


Homework Equations


pH = -log [H+]
1*10^-14 = [H+][OH-]


The Attempt at a Solution



combined volume = 49mL

n.unreacted(H+) = 0.025-0.024 = 0.001 mol

[H+] = n/V = 0.001/0.049 = 0.02 mol L-1

pH = -log[H+] = 1.69 (2 d.p)



is that correct?

many thanks,
pavadrin
 

Answers and Replies

  • #2
symbolipoint
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Yes.
 
  • #3
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okay thank you for the reply
 

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