150 m lactic acid lactic and 0100 m sodium lactate

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0.150 M lactic acid lactic and 0.100 M sodium lactate? lactic acid Ka=1.4×10-4. sodium Ci Cf HA(aq) 0.150 M 0.150 0.150 –x + - H (aq) + A (aq) 0 0.100 x 0.100 + x [ H + ][ A − ] Ka = [ HA] [ HA] [H ] = Ka [ A− ] + [A - ] pH = pK a + log [HA] pH = − log(1.4 ×10-4 ) + log[(0.100 + x) (0.150 − x)] = − log(1.4 ×10-4 ) + log[(0.100) (0.150)] = 3.854 − 0.176(Keep more digits during calculation) = 3.678=3.68(Round to significant digits at the end) Addition of strong acid or base to Buffers What What is the pH after the addition of 0.020 mole NaOH to after 0.020 1.00 L of the solution containing 0.150 M lactic acid 0.150 and 0.100 M sodium lactate? lactic acid Ka= 1.4 × 10-4 0.100 1.4 HA(aq) C’ 0.150 M Ci (0.150 M - 0.020M) Cf (0.130 – x)M + H (aq)...
View Full Document

Ask a homework question - tutors are online