For each of the solutions described below, determine the concentration of the sp
ID: 945795 • Letter: F
Question
For each of the solutions described below, determine the concentration of the species requested. Remember that pH = (-Log[H^+]) and K_w = [H^+][OH^1] = 1.0 Times 10^14 at25 degree C. If [OH^1] = 2.65 Times 10^-5 M, calculate [H^+]. If pH = 7.90, calculate [H^+]. If pH = 3.75, calculate [OH^-]. If [OH^1-] = 2.8 Times 10^-3 M, calculate pH In each reaction below, identify the acid, the base, the conjugate acid, and the conjugate base. Write the appropriate identity underneath each species. You may abbreviate conjugate acid and conjugate base as "C.A." and "C.B." respectively. CH_3CH_2COOH + NH_3 CH_3CH_2COO^1- + NH_4^1+ The equation below has a k_c value of 6.25 Times 10^9 at 300 K. If the initial concentration of SO_2 is 0.20 M and the initial concentration of O_2 is 0.10 M. determine the equilibrium concentrations of all species. 2 SO_2(g) + O_2(g) 2 SO_3(g) Now repeat the calculation if the initial concentration of both SO_2 and O_2 are 0.30 M 2 SO_2(g) + O_2(g) 2 SO_3 If [SO_2] = 0.075 M, [O_2] = 0.060 M, and [SO_2] = 2.80 M, in which direction is the reaction proceeding to reach equilibrium?Explanation / Answer
1. A [OH-] = 2.65*10-5 M
Kw = [H+][OH-]
1*10-14 = [H+]*2.65*10-5
[H+] = 1*10-14/2.65*10-5 = 3.78*10-10 M
B. PH = 7.9
-log[H+] = 7.9
[H+] = 10-7.9 = 1.25*10-8 M
C. PH = 3.75
POH = 14-PH
= 14-3.75 = 10.25
POH = 10.25
-log[OH-] = 10.25
[OH-] = 10-10.25 = 5.6*10-11 M
D. [H+] = 2.8*10-3 M
PH = -log[H+]
= -log2.8*10-3
= -log2.8+3log10
= -0.4471 +3 = 2.56
POH = 14-PH
= 14-2.56 = 11.44
2. CH3COOH + NH3 -----> CH3COO- + NH4+
acid1 Base2 conjugate base1 conjugate acid2
B. H2O + HCO3- -----> H3O+ + CO3-2
base1 acid2 conjugate acid1 conjugate base2
C. HSO4- + HPO42- -------> SO42- + H2PO4-
acid1 base2 conjugate base1 Conjugate acid2
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