{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Horvath Practice Problems 57

# Horvath Practice Problems 57 - £ C 6 H 4 NO − 2 ¤ HC 6...

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

All of the OH reacts with the H + from niacin. Now this leaves only C 6 H 4 NO 2 , but since this is the conjugate base to a weak acid we must set-up a base dissociation reaction C 6 H 4 NO 2 ­ HC 6 H 4 NO 2 + OH . The K b for this reaction is obtain from the K a for niacin and K w , K b = K w K a = 7 . 14 × 10 10 . (14.3) We will need to setup and ICE chart but fi rst we must deter- mine the concentration of C 6 H 4 NO 2 . To do this we see, 0 . 025 L HC 6 H 4 NO 2 × 0 . 10 mol HC 6 H 4 NO 2 L × 0 . 10 mol C 6 H 4 NO 2 0 . 10 mol HC 6 H 4 NO 2 = 0 . 0025 mol C 6 H 4 NO 2 But we also had 25 mL from the NaOH solution so the total volume is now 50 mL thus, £ C 6 H 4 NO 2 ¤ = 0 . 0025 mol C 6 H 4 NO 2 0 . 050 L = 0 . 05 M C 6 H 4 NO 2 Using this, the ICE chart is
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: £ C 6 H 4 NO − 2 ¤ [ HC 6 H 4 NO 2 ] £ OH − ¤ I . 05 C − x x x E . 05 − x x x . So, we have K b = 7 . 14 × 10 − 10 = ( x ) ( x ) (0 . 05 − x ) Rearranging and using the quadratic formula we get x = [ OH ] = 5 . 97 × 10 − 6 . From this pOH = − log ¡ 5 . 97 × 10 − 6 ¢ = 5 . 22 . Hence, pH = 14 − 5 . 22 = 8 . 78 . The reaction between a weak base and a strong acid would be done in a similar manner. 56 REACTIONS WITH WEAK ACIDS AND WEAK BASES...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online