The pH during the titration 30.0 mL OF 0.1000 M NaOH(aq) with 0.1000 M HCl(aq) of after 18.5 mL of the acid is added is : 12.38
First step : Determine the number of moles
Number of moles of NaOH = 0.1000 M * 30.00 mL = 3.0 mmol
Number of moles of HCl = 0.1000 M * 18.5 mL = 1.85 mmol
The net moles of NaOH = 3.0 mmol - 1.85 mmol = 1.15 mmol
Next step :
The total volume of the solution = 30.00 + 18.51 = 48.51 mL
[ NaOH ] = [ OH⁻ ] = net moles / total volume
= 1.15 mmol / 48.5 mL = 0.023711 M
Final step : Determine the pH value
pOH = – log[OH⁻] = – log ( 0.023711 )
= 1.625
Therefore :
pH value = 14 - pOH
= 14 - 1.625 ≈ 12.38
Hence we can conclude that The pH during the titration of after 18.5 mL of the acid is added is : 12.38
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