Answer:
Explanation:
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In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:
Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:
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Answer:
pH = 2.0
Explanation:
To find the pH of a solution, take the -log[H+]. In this case, the -log(9.4 x 10^-3) equals 2.02687 which makes 2.0 when accounting for significant figures.
We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
F. hold on to their protons more strongly
