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2 years ago

8

A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

f 0.235 M H2SO4 will neutralize this solution?
Please I need lots of help!!

1 answer:

igomit [66]2 years ago

8 0

<h3>Answer:</h3>

702 mL

<h3>Explanation:</h3>

We are given;

The mass of sodium hydroxide = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We are required to calculate the volume of the acid required to neutralize the solution of NaOH

<h3>Step 1: Balanced equation for the reaction</h3>

The equation for the reaction between H₂SO₄ and NaOH is given by;

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

<h3>Step 2: Calculate the number of moles of NaOH</h3>

Number of moles is given by dividing mass by molar mass

Molar mass of NaOH = 40.0 g/mol

Therefore;

Number of moles of NaOH = 13.20 g ÷ 40 g/mol

= 0.33 moles of NaOH

<h3>Step 3: Calculate the number of moles of H₂SO₄ used during the reaction </h3>

From the equation 2 moles of NaOH reacts with 1 mole H₂SO₄

Therefore, the mole ratio of NaOH : H₂SO₄ is 2 : 1

Thus, number of moles of H₂SO₄ = Moles of NaOH ÷ 2

= 0.33 moles ÷ 2

= 0.165 moles

<h3>Step 4: Calculate the volume of the H₂SO₄</h3>

Molarity refers to the concentration of a solution in moles per liter

Molarity = Number of moles ÷ Volume

Rearranging the formula;

Volume = Number of moles ÷ Molarity

= 0.165 moles ÷ 0.235 M

= 0.702 L

= 702 mL

Therefore, the volume of 0.235 M acid solution required is 702 ml

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<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
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Answer : The correct option is, (A) $[H_3O^+]=[OH^-]$ because one of each is produced every time an $[H^+]$ transfers from one water molecule to another.

Explanation :

As we know that, when the two water molecule combine to produced hydronium ion and hydroxide ion.

The balanced reaction will be:

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Acid : It is a substance that donates hydrogen ion when dissolved in water.

Base : It is a substance that accepts hydrogen ion when dissolved in water.

From this we conclude that, the hydrogen ion are transferred from one water molecule to the another water molecule to form hydronium ion and hydroxide ion. In this reaction, one water molecule will act as a base and another water molecule will act as an acid.

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The total distance traveled should not be mistaken for total displacement. While displacement measures the distance and direction from the starting position of the toy car relative to its final position, the total distance traveled is calculated by adding all the movements of the toy car together. Hence;

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A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes

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0.208 is the specific heat capacity of the metal.

Explanation:

Given:

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Heat absorbed (q) = 355 Joules

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cp (specific heat capacity) = ?

the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = $\frac{q}{mΔ T}$

c = $\frac{355}{63.5X 268.59}$

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Answer:

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Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

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