The pH after 10. 0 ml of 0. 40 M NaOH is added to 20. 0 ml of 0. 50 M HCl is 13.3 .
Calculation ,
[ [tex]OH^{-}[/tex] ] = [tex]M_{1} V_{1} -M_{2} V_{2} /V_{1} +V_{2}[/tex]
where M is molarity and V is the volume .
[ [tex]OH^{-}[/tex] ] = 0. 50 M × 20. 0 ml - 0. 40 M × 10. 0 ml / 10. 0 ml +20. 0 ml
[ [tex]OH^{-}[/tex] ] = 6/30 = 0.2
pOH = -㏒[ [tex]OH^{-}[/tex] ] = -㏒ 0.2 = 0.7
pH = 14 - pOH = 14 - 0.7 = 13.3
Therefore , the pH after 10. 0 ml of 0. 40 M NaOH is added to 20. 0 ml of 0. 50 M HCl is 13.3 .
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