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3 years ago

What volume (mL) of 7.48x10 -2 M phosphoric acid can be completely reacted with 115 mL of .244 M sodium

Chemistry

1 answer:

Y_Kistochka [10]3 years ago

8 0

Answer:125ml - volume of phosphoic acid

Explanation:

The equation of the reaction is

H3PO4 + 3 NaOH ---> Na3PO4 + 3H2O

Such that I mole of H3PO4 reacts with 3 moles of NaOH to produce the products above.

Given that

M1= Molarity of phosphoric acid = 7.48x10^-2 M

n1= number of moles of phosphoric acid = 1

V1=? ml

M2= Molarity of Sodium hydroxide =0.244 M

V2= volume of sodium hydroxide= 115ml

n2=number of moles of sodium hydroxide = 3

Using the Dilution equation formulae,

M1V1/n1 = M2V2/n2

7.48X10-2 x V1 /1= 0.244M x 115ml/3

V1=0.244Mx115mlx1/0.0748M X 3

V1= 28.06/0.2244= 125.04ml

V1=125.04ml rounded up to 125ml

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At 40.0°C, the pressure inside a nitrogen−filled tennis ball with a volume of 148 cm3 is 235 kPa. How many moles of nitrogen (N2

Musya8 [376]

To determine the number of moles of a gas, we need to have an expression that relates the pressure, temperature and volume of the system. For simplification, we assume that this gas is ideal so we use the equation PV=nRT. We calculate as follows:

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4 years ago

In the lab, frank has two solutions that contain alcohol and is mixing them with each other. he uses 100 milliliters less of sol

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Solution B (17% alcohol) = x ml
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3 years ago

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

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3 years ago