Question

ULI UF HA IS 1.0 x 10, what is the pH of the buffer? 10.00 ml of 0.200 M HCl is added to 20,0 mL of 0.0500 M NaOH, what is the pH of the solution? 1. A 100 buffer solution 240 M A LAM. Ne Calata tha alustin A

Answer 1

Molarity of HCl = 0.2 M

Volume of HCl = 10.00 mL = 0.01 L

Moles of HCl = molarity x volume in Litres = 0.2 M X 0.01 L = 0.002 moles

Molarity of NaOH = 0.05 M

Volume of NaOH = 20 mL = 0.02 L

Moles of NaOH = molarity x volume in Litres = 0.05 M X 0.02 L = 0.001 moles

Therefore,

concentration of HCl is more than concentration of NaOH.

Hence at equivalence point, concentration of H+ is higher than concentration of OH- ions.

Then,

[H+] = [moles of HCl - moles of NaOH] / total volume (volume of HCl + NaOH)

= (0.002 moles - 0.001 moles) /(0.01 L + 0.02 L)

= 0.033 M

[H+] = 0.033 M

pH = -log[H+]

= - log [0.033]

= 1.48

pH = 1.48

Therefore,

pH of given solution = 1.48