What is the [OH-] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2 ?

Question

What is the [OH-] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2 ? (A) 0.10M (B) 0.40M (C) 0.0050M (D) 0.12M

Tardigrade Admin · Accepted Answer

Number of milliequivalents of HCl =20 × 0.050 × 1=1 Number of milliequivalents of Ba ( OH )2 =2 × 30 × 0.10=6 [ OH -.] of final solution Milliequivalents of Ba ( OH )2 =(- text milliequivalents of HCl / text Total volume ) =(6-1/50)=0.1 M

Q. What is the $[O H^{-}]$ in the final solution prepared by mixing $20.0 m L$ of $0.050 M H Cl$ with $30.0 m L$ of $0.10 MB a (O H)_{2}$ ?

0.10M

0.40M

0.0050M

0.12M

Number of milliequivalents of $H Cl$
$= 20 \times 0.050 \times 1 = 1$
Number of milliequivalents of $B a (O H)_{2}$
$= 2 \times 30 \times 0.10 = 6$
$[O H^{-}$ ] of final solution
Milliequivalents of $B a (O H)_{2}$
$= \frac{- milliequivalents of H Cl}{Total volume}$
$= \frac{6 - 1}{50} = 0.1 M$

Q. What is the [OH−] in the final solution prepared by mixing 20.0mL of 0.050MHCl with 30.0mL of 0.10MBa(OH)2​ ?