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 MBa ( 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 MBa ( OH )2 ? (A) 0.40 M (B) 0.0050 M (C) 0.12 M (D) 0.10 M

Tardigrade Admin · Accepted Answer

Millmoles of H+produced =20 × 0.05=1 Millmoles of O H-produced =30 × 0.1 × 2=6 (Each Ba ( OH )2 gives 2 OH -). Millmoles of O H- remaining in solution =6-1=5 Total volume of solution =20+30=50 mL [O H-]=(5/50)=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 $0.10 MB a (O H)_{2} ?$

0.40 M

0.0050 M

0.12 M

0.10 M

Millmoles of $H^{+}$ produced $= 20 \times 0.05 = 1$
Millmoles of $O H^{-}$ produced $= 30 \times 0.1 \times 2 = 6$
(Each $B a (O H)_{2}$ gives $2 O H^{-}$ ).
Millmoles of $O H^{-}$ remaining in solution $= 6 - 1 = 5$
Total volume of solution $= 20 + 30 = 50 m L$
$[O H^{-}] = \frac{5}{50} = 0.1 M$

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​?