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arsen [322]
3 years ago
12

A lake in Utah has a hydronium ion (H3O+) concentration of 4.60 x 10-9 M. Calculate the pH of the water.

Chemistry
1 answer:
klio [65]3 years ago
8 0
Answer is: <span>pH of the water is 8.33, this is basic solution (pH is greater than seven).

</span>[H₃O+] = 4.60·10⁻⁹ M; equilibrium concentration of hydronium ion.
pH = -log[H₃<span>O+].
</span>pH = -log(4.60·10⁻⁹ M).
pH = 8.33.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity <span>an aqueous solution.
</span><span>
</span>
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To prepare an acetic acid/acetate buffer, a technician mixes 30.6 mL of 0.0880 acetic acid and 21.6 mL of 0.110 sodium acetate i
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Answer: There are 0.00269 moles of acetic acid in buffer.

Explanation:

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\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}     .....(1)

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Volume of solution = 30.6 mL

Putting values in equation 1, we get:

0.0880M=\frac{\text{Moles of acetic acid}\times 1000}{30.6ml}\\\\\text{Moles of acetic acid}=\frac{0.0880\times 30.6}{1000}=0.00269mol

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2 years ago
In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O
yarga [219]
HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

K_{a} = \frac{[ H_{3}  O^{+} ][O Cl^{-} ]}{HClO}

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
      
          HClO       + H2O   ⇄   H3O+ +  OCl-
I     1.0x10^-4                          0             0
C        -x                                 +x           +x 
E  (1.0x10^-4 - x)                     x             x

Substituting the excess (E) concentration to the Ka equation:

K_{a} = \frac{[x ][x]}{1.0 \ x \  10^{-4} - x }

Simplifying the equation would yield a quadratic equation:

x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
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Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
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Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-


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