Equivalence HCL(mL) Equivalence HCL(L) Temp (degC) Temp (K) 1/T(K^-1) [Ca] [OH-]
ID: 841041 • Letter: E
Question
Equivalence HCL(mL)
Equivalence HCL(L)
Temp (degC)
Temp (K)
1/T(K^-1)
[Ca]
[OH-]
Ksp
In(Ksp)
8.65
0.00865
21
294.15
0.00339963
0.00973774
0.01947548
3.69347E-06
-12.508945
6.69
0.00669
65
338.15
0.00295727
0.00753127
0.01506254
1.70869E-06
-13.279781
6.43
0.00643
75
348.15
0.00287233
0.00723857
0.01447715
1.51712E-06
-13.398699
From the Data above, Show the full calculation for determination of Ksp for one of the
temperature trials, 21 degrees celcius. Also, calculate DeltaH and show work.
[HCl] = 0.04503 M
Thank you!
Equivalence HCL(mL)
Equivalence HCL(L)
Temp (degC)
Temp (K)
1/T(K^-1)
[Ca]
[OH-]
Ksp
In(Ksp)
8.65
0.00865
21
294.15
0.00339963
0.00973774
0.01947548
3.69347E-06
-12.508945
6.69
0.00669
65
338.15
0.00295727
0.00753127
0.01506254
1.70869E-06
-13.279781
6.43
0.00643
75
348.15
0.00287233
0.00723857
0.01447715
1.51712E-06
-13.398699
Explanation / Answer
1) Determine moles of HCl used:
moles HCl = (0.04503 mol / L) (0.00865 L) = 0.0003895 mol =3.895 * 10-4 mol
2) Determine moles Ca(OH)2 titrated:
0.0003895 mol / 2 = 0.0001947 mol
Remember, every one Ca(OH)2 titrated requires 2 H+
3) Convert moles Ca(OH)2 to grams Ca(OH)2:
0.0001947 mol times 74.0918 g/mol = 0.01443 g
Let volume of Ca(OH)2 titrated be V (in L).
This is grams per V (in L)
1) 0.0001947 mol of Ca(OH)2 in V (in L) means:
[Ca2+] = 0.0001947 mol / V (in L) = 0.0097377 M = > V = 0.02 L = 20 mL
[OH
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