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3 years ago

What is the pH of a solution that has [H+] of 1.659e^-9 M

2 answers:

8 0

PH of a solution is negative log of its hydrogen ion concentration. Mathematically it can be written as:
$pH = - log[ H^{+}]$
For the said problem pH will be:
$pH = -log [ 1.659 * 10^{-9} ] \\ pH=-(log[1.659]+log[10^{-9}]) \\ pH=-log[1.659]+9log[10] \\ pH=-0.22+9=8.78$

Thus the pH of solution will be 8.78.

ella [17]3 years ago

4 0

Answer is: pH = 8,75.
c(H⁺ or H₃O⁺) = 1,659·10⁻⁹ M = 1,659·10⁻⁹ mol/L = 0,000000001659 mol/L.

pH = -logc(H₃O⁺).

pH = -log(1,659·10⁻⁹ mol/L).

pH = 8,75.
When pH is less than seven (pH<7), solution is acidic.
When is equal seven (pH = 7), solution is neutral.
When pH is greater than seven (pH > 7), solution is basic (like this example).

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What is the molarity (molar concentration, unit = M) of K+ found in 200 mL 0.2 M K2HPO4 solution?

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Answer:

0.4 M

Explanation:

The process that takes place in an aqueous K₂HPO₄ solution is:

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200 mL * 0.2 M = 40 mmol K₂HPO₄

Then we convert K₂HPO₄ moles into K⁺ moles, using the stoichiometric coefficients of the reaction above:

40 mmol K₂HPO₄ * $\frac{2mmolK^+}{1mmolK_2HPO_4}$ = 80 mmol K⁺

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3 years ago

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The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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When the equation BiCl3 + H2S → Bi2S3 + HCl is the coefficient of HCl will be

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The product Bismuth (III) sulfide ( $Bi_{2}S_{3}$ ) is used as a starting material for the synthesis of other bismuth compounds. It is formed, together with an acid, from a reaction between a Bismuth (III) salt and hydrogen sulfide ( $H_{2}S$ ). In the given reaction, the Bismuth salt used in the reaction is Bismuth (III) chloride ( $BiCl_{3}$ ).

The balanced reaction of $BiCl_{3}$ and $H_{2}S$ is

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