Chemistry, 31.03.2020 04:25 VamPL

5. Phosphoric acid (H3PO4) is a triprotic acid with three ionizable protons. Write a balance equation for the neutralization of phosphoric acid with NaOH. How many milliliters of 0.120 M NaOH would be required to completely neutralize 35.0 ml of 0.0440 M H3PO4

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Answer from: benchang3621

1. H3PO4 + 3NaOH —> Na3PO4 + 3H2O

2. 38.5mL

Explanation:

1. We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

2. H3PO4 + 3NaOH —> Na3PO4 + 3H2O

From the equation above, the following data were obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 3

Data obtained from the question include:

Vb (volume of base) =?

Mb (Molarity of base) = 0.120 M

Va (volume of acid) = 35.0 mL

Ma (Molarity of acid) = 0.0440 M

Using the formula MaVa/MbVb = nA/nB, the volume of the base (i.e NaOH) can be obtained as follow:

MaVa/MbVb = nA/nB

0.0440 x 35/ 0.120 x Vb = 1/3

Cross multiply to express in linear form as shown below:

0.120 x Vb = 0.0440 x 35 x 3

Divide both side by 0.120

Vb = (0.0440 x 35 x 3) /0.120

Vb = 38.5mL

Therefore, 38.5mL of 0.120 M NaOH is needed for the complete neutralization.

Answer from: jasmineeeee449292

We need 38.5 mL of NaOH to neutralize the H3PO4 solution

Explanation:

Step 1: Data given

Molarity of NaOH = 0.120 M

Volume of H3PO4 = 35.0 mL = 0.035 L

Molarity of H3PO4 = 0.0440 M

Step 2: The balanced equation

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

Step 3: Calculate the volume of NaOH

b*Ca*Va = a *Cb*Vb

⇒with b = the coefficient of NaOH = 3

⇒with Ca = the concentration of H3PO4 = 0.0440 M

⇒with Va = the volume of H3PO4 = 35.0 mL = 0.0350 L

⇒with a = the coefficient of H3PO4 = 1

⇒with Cb = the concentration of NaOH = 0.120 M

⇒with Vb = the volume of NaOH = TO BE DETERMINED

3*0.0440 * 0.0350 = 0.120 * Vb

Vb = 0.0385 L = 38.5 mL

We need 38.5 mL of NaOH to neutralize the H3PO4 solution

Answer from: Quest

Bthats what im doing to and i figured it out it is b

Answer from: Quest

the formula for the conversion of temperature from degree celsius to kelvin is:

$k = ^{o}c + 273.15$

the sample of water at $38^{o}c$ is converted to kelvin temperature by using the above formula as:

$k = 38^{o}c + 273.15 = 311.15 k$

on comparing the two samples of water at $38^{o}c$ or we can say that at $311.15 k$ and at $295 k$ respectively. the difference between the temperatures of two samples of water is:

$311.15 k - 295 k = 16.15 k$

hence, one can say that the sample of water at $38^{o}c$ is greater than the sample of water at $295 k$ by $16.15 k$ .

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5. Phosphoric acid (H3PO4) is a triprotic acid with three ionizable protons. Write a balance equatio...

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