4 ml of HCl solution of pH =2 is mixed with 6 ml of NaOH solution of pH =12. What would be the final pH of solution ? log 2=0.3

Question

4 ml of HCl solution of pH =2 is mixed with 6 ml of NaOH solution of pH =12. What would be the final pH of solution ? log 2=0.3 (A) 10.3 (B) 11.3 (C) 11 (D) 4.3

Tardigrade Admin · Accepted Answer

pH =2,[ HCl ]=10-2( M ) pOH =2 or [ NaOH ]=10-2( M ) 4 ml of 10-2 (M) HCl ≡ 4 × 10-5 moles HCl. 6 ml of 10-2 (M) NaOH ≡ 6 × 10-5 moles NaOH After mixing excess moles of OH -=2 × 10-5 [ OH -]=(2 × 10-5/10) × 103=2 × 10-3 or pOH =3- log 2=3-0.3=2.7 or pH =11.3

Q. 4ml of HCl solution of pH=2 is mixed with 6ml of NaOH solution of pH=12. What would be the final pH of solution ?log2=0.3

10.3

11.3

11

4.3

pH=2,[HCl]=10−2(M)&pOH=2 or [NaOH]=10−2(M) 4ml of 10−2 (M) HCl≡4×10−5 moles HCl. 6ml of 10−2 (M) NaOH≡6×10−5 moles NaOH After mixing excess moles of OH−=2×10−5 [OH−]=102×10−5​×103=2×10−3 or pOH=3−log2=3−0.3=2.7 or pH=11.3

Q. $4 m l$ of $H Cl$ solution of $p H = 2$ is mixed with $6 m l$ of $N a O H$ solution of $p H = 12$ . What would be the final $p H$ of solution $? lo g 2 = 0.3$