Chemistry, 23.03.2020 22:59 kev71

A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of solution and was then titrated with 0.140 M . After 500.0 mL was added, the pH was 4.63. The experimenter found that 1.00 L of 0.140 M was required to reach the stoichiometric point of the titration. a What is the molar mass of ? Molar mass = 129.14 g/mol b Calculate the pH of the solution at the stoichiometric point of the titration.

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Answer from: mricee9718

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

$V_{equivalence}_{acid}$ = 1.00 L

Equation for the reaction can be represented as:

$NaA_{(aq)} + HCl_{(aq)} ----- HA_{(aq)} + NaCl_{(aq)$

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = $\frac{18.08g}{0.140 mole}$

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = $10^{-4.63]$

= $2.3*10^{-5}$

$[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}$

$= \frac{0.140 mol}{1.116 L}$

= 0.1255 M

K_a of HA = $2.3*10^{-5}$

$K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}$

$= 4.35*10^{-10}$

$A_{(aq)}$ + $H_2O_{(l)}$ $\rightleftharpoons$ $HA_{(aq)}$ + $OH^-_{(aq)}$

Initial 0.1255 0 0

Change - x + x + x

Equilibrium 0.1255 - x x x

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}$

As K_b is very small, (o.1255 - x) = 0.1255

$x = \sqrt{0.1255*4.35*10^{-10}}$

[OH⁻] = $x = 7.4 *10^{-6}$

But pOH = - log [OH⁻]

= - log [ $7.4*10^{-6}$ ]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

Answer from: Quest

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"

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A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough...