1
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?
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?
Open in App
Solution
The correct option is A 0.10 M
Given, VHCl=20 mL, VBa(OH)2=30 mL, [HCl]=0.050 M, [Ba(OH)2]=0.10 M
mmole=N×V(in mL)
mmole=M×valence factor×V(in mL)
mmole of HCl=20×0.05=1
mmole of Ba(OH)2=30×0.1×2=6
∵Ba(OH)2⇌Ba2++2 OH−
∴ mmol of Ba(OH)2 left in mixture=6−1=5
Total volume =20+30=50 mL
NBa(OH)2 or [OH−]=550=0.1 M
Given, VHCl=20 mL, VBa(OH)2=30 mL, [HCl]=0.050 M, [Ba(OH)2]=0.10 M
mmole=N×V(in mL)
mmole=M×valence factor×V(in mL)
mmole of HCl=20×0.05=1
mmole of Ba(OH)2=30×0.1×2=6
∵Ba(OH)2⇌Ba2++2 OH−
∴ mmol of Ba(OH)2 left in mixture=6−1=5
Total volume =20+30=50 mL
NBa(OH)2 or [OH−]=550=0.1 M
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
