The pH of a solution prepared by mixing 50.0 mL of 0.125 M NaOH and 40.0 mL of 0.125 M HNO3 [Solved]

pH = 14.00 pOH
pH = 14.00 1.86
pH = 12.14

pOH = -log( OH^(-) )
pOH = -log( 0.0139 )
pOH = 1.86

0.00500 mol OH^(-) will neutralize 0.00500 mol H^( ) so were just left with 0.00125 mol OH^(-) ( Im now guessing thats where your 0.00125 came from ) so now just find the molarity of OH^(-)

0.125M HNO3 * 0.0400L = 0.00500 mol H^( )
0.125M NaOH * 0.0500L = 0.00625 mol OH^(-)

0.00500 mol OH^(-) will neutralize 0.00500 mol H^( ) so were just left with 0.00125 mol OH^(-) ( Im now guessing thats where your 0.00125 came from ) so now just find the molarity of OH^(-)

pH = 14.00 pOH
pH = 14.00 1.86
pH = 12.14

pOH = -log( OH^(-) )
pOH = -log( 0.0139 )
pOH = 1.86

pH = 14.00 pOH
pH = 14.00 1.86
pH = 12.14

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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