All categories

2 years ago

Which compound releases hydroxide ions in an aqueous solution?

Chemistry

2 answers:

myrzilka [38]2 years ago

6 0

Answer:

The correct answer is option 4.

Explanation:

Substance which releases hydroxide ions( $OH^-$ ) in an aqueous solution are termed as bases.

Substance which releases proton ( $H^+$ ) in an aqueous solution are termed as acids.

From the given options:

$CH_3COOH$ : It is an weak organic acid known as acetic acid. Being acid it releases $[H&+]$ ion in an aqueous solution.

$CH_3COOH(aq)\rightarrow CH_3COO^-(aq)+H^+(aq)$

$CH_3COH$ : It is an alcohol named methanol and it is slightly acidic.They gives $H^+$ ions and alkoxiode ions.

$CH_3COH(aq)\rightarrow CH_3CO^-(aq)+H^+(aq)$

$HCl$ : It is strong acid known as hydrochloric acid. Being highly acidic it releases $[H&+]$ ion in an aqueous solution.

$CH_3COOH(aq)\rightarrow Cl^-(aq)+H^+(aq)$

$KOH$ : It is strong base known as potassium hydroxide. Being highly acidic it releases $[H&+]$ ion in an aqueous solution.

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

chubhunter [2.5K]2 years ago

5 0

4- KOH

KOH is a strong base. In water it dissociates as
<span>KOH--> K+ + OH- </span>
<span>A is acetic acid, a weak acid </span>
<span>CH3OH is an alcohol, weakly acidic in water </span>
<span>HCl is a strong acid.</span>

You might be interested in

A 103.8g sample of nitric acid solution that is 70.0% HNO3 contains by mass

soldi70 [24.7K]

w/w percentage <span>
= mass of the pure compound / total mass of the sample x 100%

70% HNO₃ contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>

The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
= 103.8 g x (70 / 100)
<span> = 72.66 g = 72.7 g.</span></span>

3 0

2 years ago

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

2 years ago

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

3 0

2 years ago

100 points!!!!!!!!!!!!!!!!!! how does a volcanic eruption benefit the surrounding area? A. Lava and ash bury animal habitats. B.

Shkiper50 [21]

I think it would be C) The surrounding soil can become very fertile

4 0

2 years ago

Read 2 more answers

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp

weeeeeb [17]

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$ .

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.

5 0

2 years ago

Read 2 more answers