A 30.0 ml sample of 0.200 M Ba(OH)2 is titrated with 0.100 M HCl. teh pH will be:
a) The pH of the solution before the addition of HCl is 13.301.
b) The pH of the solution after the addition of 25.0 mL HCl is 13.494.
pH is potential hydrogen. It is the measurement of the acidity and basicity of any solution.
a)
[tex]\rm Ba(OH)_2 = Ba_2+ + 2OH^-Ba(OH)_2[/tex]
Given, [OH¯] = 0.200 M
pOH = -log 0.200 = 0.699
pH = 14.000 - pOH = 14.000 - 0.699 = 13.301
b)
Now 25 ml HCl is added
0.025 × 0.100 = 0.0025
Moles of OH = 0.02 × 0.0025 = 0.0005
Remaining OH = 0.0025 – 0.0005 = 0.00245
Now calculate the pH
0.055 is divided by 0.200 = 0.275
pOH = -log 0.275 = 0.5606
pH = 14.000 - 0.5606 = 13.4394
Thus, a) 13.301.
b) 13.4394.
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