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# pH at the equilibrium point

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6) When 12.0ml of 0.250 M NaOH is added to 10.0ml of 0310 M HCL, has equilibrium been reached? Explain.
What is the pH at this point ?

https://brainmass.com/chemistry/acids-and-bases/ph-equilibrium-point-173733

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acid base 6
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6) Strong acid - strong base
When 12.0ml of 0.250 M NaOH is added to 10.0ml of 0310 M HCL , has equilibrium
been reached ? EXPLAIN
What is the pH at this point ?

First write down the chemical reaction equation:
For the strong acid and strong base, they completely ionized in the solution.
NaOH ïƒ  Na+ + OH-
HCl ïƒ  Cl- + H+
So
NaOH + HCl ïƒ  NaCl + H2O

12.0 ml of 0.250 M NaOH is 12.0 * 0.250 = 3 mmol of NaOH
10.0 ml of 0.310 M HCl is 10.0 * 0.310 = 3.1 mmol of HCl.
From the above reaction, 1 mol of NaOH reacts with 1 mol of HCl to reach the equilibrium since NaCl is a neutral salt.

Therefore, 3 mmol of NaOH reacts with 3 mmol of HCl. And 3.1 - 3 = 0.1 mmol HCl is left.
NaOH + HCl ïƒ  NaCl + H2O
Beginning 3 mmol 3.1 mmol
Equilibrium 0 0.1 mmol 3 mmol

So the equilibrium has not been reached.
There is 0.1 mmol HCl in the solution.
The total volume is 12.0 + 10.0 = 22.0 ml.
So the concentration of H+ is
x = 0.1 mmol / 22.0 ml = 0.004545 M.
the pH of the solution is
pH = -log[H+] = -logx = -log 0.004545 = 2.34.

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