Answer: -
1.8
End point passed.
Explanation: -
Volume of HI solution = 47.0 mL = 0.047 L
Strength of HI solution = 0.47 M
Since HI is a strong acid, all of HI will dissociate to give H +.
[H+ ] =0.47 M x 0.047 L
= 0.02209 mol
Volume of KOH = 25.0 mL = 0.025 L
Strength of KOH = 0.25 M
Since KOH is a strong base, all of KOH will dissociate to give OH-.
[OH-] = 0.25 M x 0.025L
= 0.00625 mol
Since [H+] and [OH-] react to form water,
[H+] unreacted = 0.02209 – 0.00625 = 0.01584 mol
Using the formula
pH = - log [H+]
= - log 0.01584
= 1.8
As the strong acid HI is being titrated by strong base KOH, the pH at the end point should be 7.
The pH has already crossed that. Thus the titration end point has already passed
After titration, there are more moles of HI than KOH, implying excess HI (acid) is present. The remaining acid concentration is 0.2 M and consequently, the final pH of the solution is approximately 0.70.
In the case of the titration of a 47.0 mL of 0.47 M HI solution with 25.0 mL of 0.25 M KOH, we first need to understand that HI is a strong acid and KOH is a strong base. When we titrate a strong acid with a strong base, the equivalence point occurs at a pH of 7.0.
First, we calculate the moles of the acid and the base: moles of HI = 0.47 mol/L * 0.047 L = 0.02209 mol, and moles of KOH = 0.25 mol/L * 0.025 L = 0.00625 mol. Since there are more moles of HI than KOH, we will have extra HI left after the titration. Hence, it is a strong acid-strong base titration before the equivalence point i.e. when we have excess acid.
The remaining acid concentration is (0.02209 mol - 0.00625 mol) / (0.047 L + 0.025 L) = 0.2 M and pH of a strong acid is basically the negative logarithm of the acid's concentration. Therefore, the pH is -log[H+] = -log(0.2) = approx. 0.70.
#SPJ3
b. molality.
c. parts per million.
Answer:
Decreases abs basicity decrease
Explanation: