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