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Advocard [28]
3 years ago
6

What is the pH of a solution of KOH with a hydroxide concentration of [OH⁻] = 1.10 x 10⁻⁴

Chemistry
1 answer:
olga2289 [7]3 years ago
3 0

Answer:

p[H+] = 10.042

Explanation:

As we know that

pKw = pH + pOH......eq (1)

we will calculate the pH of OH- and then we will calculate the pH of H+

So p[OH-] = - log [1.10 * 10^{-4}]

Solving the right side of the equation, we get

p[OH-]

= - [-3.958]\\= 3.958

Now we know that

pKw = 14.0

Substituting the value of pOH in the above equation, we get -

14.0 = p[H+] + 3.958\\p[H+] = 14 - 3.958\\p[H+] = 10.042

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Answer:

A) oxidizing agent is SO2

B) NaClO is the oxidizing agent

Explanation:

A) This is a redox reaction in which oxidation and reduction occur simultaneously.

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B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)

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What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n
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<u>Answer:</u> The volume of permanganate ion (potassium permanganate) is 10.0 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

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Putting values in equation 1, we get:

0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol

For the given chemical equation:

5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = \frac{1}{5}\times 0.00762=0.00152mol of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL

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