Ask question

Ask question

All categories

3 years ago

6

A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution? 1.0 × 10–5 M OH– 1.0 ×

10–14 M OH– 1.0 × 105 M OH– 1.0 × 10–9 M OH–

2 answers:

Triss [41]3 years ago

6 0

Answer:

1.0 × 10⁻⁹ M.

Explanation:

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.</em>

[H₃O⁺] = 1.0 x 10⁻⁵ M.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(1.0 x 10⁻⁵ M) = <em>1.0 × 10⁻⁹ M.</em>

Artyom0805 [142]3 years ago

4 0

Answer : The correct option is, $1.0\times 10^{-9}M$ $OH^-$

Explanation:

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-5})$

$pH=5$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-5=9$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$9=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-9}M$

Therefore, the $OH^-$ concentration is, $1.0\times 10^{-9}M$

You might be interested in

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

What do you know about a chemical compound by looking at its chemical formula?

coldgirl [10]

The type of bonds present in the compound. and the type of structure it has and the elements that are presents and the number of moles of each element in one mole of the compound.

4 0

3 years ago

Read 2 more answers

The density of gold is 19.3 g /cm cubed The density of iron pyrite is 5.0 g /cm cubed. Is a nugget of iron pyrite and a nugget o

telo118 [61]

If you clear volume in the density equation:

$\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}$

The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.

$V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3$

$V_{iron} = \frac{m}{\rho} = \frac{50\ g}{5.0\ g/cm^3} = \bf 10\ cm^3$

6 0

3 years ago

What is the pH of a sample with [OH-] = 2.5 x 10-11 M?

sveticcg [70]

Answer:

The pH of the sample is 3,4.

Explanation:

We calculate the pOH from the formula pOH = -log (OH-). We know that for all aqueous solutions: pH + pOH = 14, and from there we clear pH:

pOH= -log (OH-)=10,60

pH + pOH = 14

pH + 10,60 = 14

pH=14 -10,60

<em>pH=3,4</em>

3 0

3 years ago

Lithium is a metal.

Zina [86]

Answer:

Explanation:

lithium: lithium is very soft, silvery metal. melting point is 180.54°C and boiling point is 1,335°C. it's density is 0.534 grams per cu.cm. oxygen: oxygen is colourless , odorless , tasteless gas

8 0

3 years ago