Someone help me please- Im stuck, will give brainliest

A 25.00-ml sample of propionic acid, hc3h5o2, of unknown concentration was titrated with 0.141 m koh. the equivalence point was

Scorpion4ik [409]

The concentration of propionate ions at equivalent point is 0. 084 M
calculation
write the equation for reaction
HC3H5O2 + KOH → H2O + KC3H5O2

find the moles of KOH used
moles= molarity × volume in liters
volume in liters = 37.48 ml/1000 =0.03718L
molarity = 0.141 M

moles is therefore = 0.141 M x0.03718 = 5.24238 x10^-3 moles

by use of mole ratio between KOH and K3C3H5O2 which is 1:1 the moles of K3C3H5O2 is also = 5.24238 x10^-3 moles

molarity is therefore = number of moles/volume in liters
volume in liters = 25/1000 +37.48/1000=0.06248 L

molarity of propionate ions is therefore = (5.24238 x10^-3) / 0.06248 = 0.084 M

7 0

3 years ago

The NH₄⁺ ion forms acidic solutions, and the CH₃COO⁻ ion forms basic solutions. However, a solution of ammonium acetate is almos

tigry1 [53]

Not always ammonium salts of weak acids form neutral solutions.

When formic acid reacts with ammonia, ammonium formate is produced:

HCO2H + NH3 ----> NH4HCO2

You already know that the weak conjugate bases of NH3 and HCO2H are NH4+ and HCO2, respectively.

How can the pH of the solution be calculated if the salt's anion causes the pH to rise and the salt's cation causes it to fall? The relative intensities of the basic anion and the acidic cation hold the key to the solution.

As was already established, formate is a weak base and will create hydroxide ions in water, whereas ammonium is a weak acid and will make hydronium ions in water.

NH4⁺ + H2O -----> NH3 + H3O⁺

HCO2⁻ + H2O -----> HCO2H + OH⁻

Since the acid ionization of NH4+ is more favored than the base ionization of HCO2-, the solution will be acidic.

To learn more about ammonium salts:

brainly.com/question/10874844

#SPJ4

6 0

1 year ago

How many molecules are in 23.47 grams of CO2?

Genrish500 [490]

Do a quick conversion: 1 grams Co2 = 0.0084841820909097 mole using the molecular weight calculator and the molar mass of Co2.

8 0

3 years ago

Ion Concentration 1. Which solution has the greatest [SO42-]: a) 0.075 M H2SO4 b) 0.15 M Na2SO4 c) 0.080 M Al2(SO4)3?

soldi70 [24.7K]

Answer:

c) 0.080 M Al₂(SO₄)₃

Explanation:

Ion [SO₄²⁻] concentration of each solution is:

a) 0.075 M H₂SO₄: <em>[SO₄²⁻] = 0.075M</em>. Because 1 mole of H₂SO₄ contains 1 mole of SO₄²⁻

b) 0.15 M Na₂SO₄: <em>[SO₄²⁻] = 0.15M</em>. Also, 1 mole of Na₂SO₄ contains 1 mole of SO₄²⁻

c) 0.080 M Al₂(SO₄)₃ [SO₄²⁻] = 0.080Mₓ3 =<em> 0.240M</em>. Because 1 mole of Al₂(SO₄)₃ contains 3 moles of SO₄²⁻.

<h3>Thus, the soluion that has the greatest [SO₄²⁻] is 0.080 M Al₂(SO₄)₃</h3>

8 0

3 years ago

The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra

Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

Br2(g) ⇌ 2Br(g)

Initially 0.0345M 0.0416M

$Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016$

$Q_C =0.05016 >>> K_c(0.00584)$

Thus, the given reaction will proceed in the backward direction

The I.C.E table is as follows:

Br2(g) ⇌ 2Br(g)

I 0.0345 0.0416

C +x -2x

E (0.0345+x) (0.0416 -2x)

$K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584$

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908 - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] = (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] = (0.0416 -2(0.0126)) = 0.0164 M

4 0

3 years ago