Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka=1.3x10-5) after...

Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; K_a=1.3x10^-5) after adding 22.18 mL of 0.1000M NaOH.

** All volumes should have a minimum of 2 decimal places

Solutions

Expert Solution

Given 40 mL of 0.1 M CH3CH2COOH and 22.18 mL of 0.1 M NaOH

Moles of CH3CH2COOH = 0.04 L x 0.1 = 0.004 moles

Moles of NaOH = 0.02218 L x 0.1 = 0.002218 moles

NaOH + CH3CH2COOH -------- H20 + NaCH3CH2COO

0.002218 0.004 0 initially

0 0.004-0.002218 0.002218 finally

Now pH = pKa + log [NaCH3CH2COO] / [CH3CH2COOH ]

given Ka =1.3 x 10^-5 from this pKa = - log Ka = 4.886

Therefore pH = 4.886 + log 0.002218/ 0.001782 = 4.886 + 0.095 = 4.98 which is acidic solution

queen_honey_blossom answered 1 year ago

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