500.0mL of 0.120M NaOH is added to 535mL of 0.250M weak acid (Ka= 5.88*10^-5) What is...

500.0mL of 0.120M NaOH is added to 535mL of 0.250M weak acid (Ka= 5.88*10^-5) What is the pH of the resulting buffer?

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Expert Solution

No. of moles of NaOH present = molarity bx volume in liters = 0.5L x 0.120= 0.06moles NaOH

No. of moles of acid present = 0.535L x 0.250 = 0.13375 moles

HX + NaOH ----> NaX + H2O

0.13375 0.06 - -

-0.06 -0.06 0.06 -

0.07375 0 0.06 -

Noe, pH = pKa + log[A-]/[HA]

Given, Ka = 5.88E-5 so, pKa = -log[Ka] = 4.23

pH = 4.23 + log(0.06/0.07375) = 4.14

queen_honey_blossom answered 8 months ago

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