Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to 35.0 mL of 0.250 M NaOH(aq).
The following solution is suggested to handle the subject “Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to 35.0 mL of 0.250 M NaOH(aq).“. Let’s keep an eye on the content below!
Question “Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to 35.0 mL of 0.250 M NaOH(aq).”
Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to 35.0 mL of 0.250 M NaOH(aq).
Calculate the pH of the resulting solution if 25.0 mL
of 0.250 M HCl(aq) is added to 15.0 mL of 0.350 M NaOH(aq).
No referals as the only answer unless you’re the first one.
Answer
1)
Given:
M(HCl = 0.25M
V(HCl), 25 mL
M(NaOH = 0.25M
V(NaOH = 35 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl), = 0.25 M* 25 mL = 6.25 mg
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH). = 0.25 M* 35 mL = 8.75 mgmol
We have:
6.25 mol(HCl).
mol(NaOH = 8.75mmol
6.25 mmol each will react
Resting mol of NaOH =2.5 mmol
Total volume = 60.0 mL
[OH-] = Mol of Base Remaining / Volume
[OH-] = 2.5 mg/60.0 mL
4.167*10-2M
use:
pOH = -log [OH-]
= -log (4.167*10^-2)
= 1.3802
use:
PH = 14. – pOH
= 14 – 1.3802
= 12.6198
Answer: 12.62
2)
Given:
M(HCl = 0.25M
V(HCl), 25 mL
M(NaOH = 0.35M
V(NaOH = 15 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl), = 0.25 M* 25 mL = 6.25 mg
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH), = 0.35 M * 15. mL = 5.25mmol
We have:
6.25 mol(HCl).
mol(NaOH = 5.25mmol
5.25 mmol each will react
Resting mol of HCl = 1.0 mmol
Total volume = 40.0 mL
[H+] = Mol of remaining acid / volume
[H+] = 1 mg/40.0 mL
= 2.5*10-2M
use:
pH = -log [H+]
= -log (2.5*10^-2)
= 1.6021
Answer: 1.60
Conclusion
Above is the solution for “Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to 35.0 mL of 0.250 M NaOH(aq).“. We hope that you find a good answer and gain the knowledge about this topic of Ph.