When 80 mL of 0.20 M HCl is mixed with 120 mL of 0.15 M KOH, the resultant solution is the same as a solution of

Question

When 80 mL of 0.20 M HCl is mixed with 120 mL of 0.15 M KOH, the resultant solution is the same as a solution of (A) 0.16 M KCl and 0.02 M HCl (B) 0.08 M KCl (C) 0.08 M KCl and 0.01 M KOH (D) 0.08 M KCl and 0.01 M HCl

Tardigrade Admin · Accepted Answer

80 mL of 0.20 M HCl =80 × 0.2 =16 millimol HCl 120 mL of 0.15 M KOH =120 × 0.15 = 18 millimol KOH M1 V1( HCl ) < M2 V2( KOH ) Thus, resulting solution is basic containing KCl and unreacted KOH KOH + HCl arrow KCl + H 2 O KOH =(M2 V2-M1 V1/V1+V2)=(18-16/200)=0.01 M KCl formed = HCl used =16 millimoles =16 × 10-3 mol in 200 mL or 0.2 L solution [ KCl ]=(16 × 10-3/0.200 L )=0.08 M

Q. When $80 m L$ of $0.20 M H Cl$ is mixed with $120 m L$ of $0.15 M K O H$ , the resultant solution is the same as a solution of

$0.16 M K Cl$ and $0.02 M H Cl$

$0.08 M K Cl$

$0.08 M K Cl$ and $0.01 M K O H$

$0.08 M K Cl$ and $0.01 M H Cl$

Q. When 80mL of 0.20MHCl is mixed with 120mL of 0.15MKOH, the resultant solution is the same as a solution of

0.16MKCl and 0.02MHCl

0.08MKCl

0.08MKCl and 0.01MKOH

0.08MKCl and 0.01MHCl

Q. When $80 m L$ of $0.20 M H Cl$ is mixed with $120 m L$ of $0.15 M K O H$ , the resultant solution is the same as a solution of

$0.16 M K Cl$ and $0.02 M H Cl$

$0.08 M K Cl$

$0.08 M K Cl$ and $0.01 M K O H$

$0.08 M K Cl$ and $0.01 M H Cl$