What is the concentration of an HCl solution if 23.1 mL of HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2
I cannot get the right answer
If you had shown your work I could have found the error.
(Macic)(23.1mL)=(0.101M)(35.0mL)
Macid = .153
See your post above.
To determine the concentration of the HCl solution, you need to use the balanced chemical equation for the neutralization reaction between HCl and Ba(OH)2:
2HCl + Ba(OH)2 → 2H2O + BaCl2
From the equation, you can see that it takes two moles of HCl to neutralize one mole of Ba(OH)2.
First, determine the number of moles of Ba(OH)2 using its concentration and volume:
moles of Ba(OH)2 = concentration × volume
= 0.101 M × 35.0 mL
= 0.00354 moles
Based on the balanced equation, the number of moles of HCl is twice that of Ba(OH)2, so the number of moles of HCl is:
moles of HCl = 2 × moles of Ba(OH)2
= 2 × 0.00354 moles
= 0.00708 moles
Finally, calculate the concentration of the HCl solution using the number of moles and volume:
concentration of HCl = moles of HCl / volume of HCl solution
= 0.00708 moles / 23.1 mL
= 0.306 M
Therefore, the concentration of the HCl solution is 0.306 M.