Calculate the pH during the titration of 30.00 mL of 0.1000 M NaOH(aq) with 0.1000 M HCl(aq) after 18.51 mL of the acid have bee

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zlopas [31] · Answer 1 · 2023-03-02T21:47:52+03:00

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

<h3 /><h3>Calculate the pH value during the titration </h3>

<u>First step</u> : 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

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

<u>Final step</u> : 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