The charge on a magnesium ion is +2. When it formed an ionic compound with another element, it lost 2... what?

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0

3 years ago

What is the mass of 4.99×1021 platinum atoms?

charle [14.2K]

Answer:

$\boxed {\boxed {\sf 1.62 \ g \ Pt}}$

Explanation:

We are asked to find the mass of a number of platinum (Pt) atoms.

<h3>1. Convert Atoms to Moles </h3>

First, we convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are platinum atoms.

We will convert using dimensional analysis so we set up a ratio using the information we know (6.022 × 10²³ platinum atoms in 1 mole of platinum).

$\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

We are converting 4.99 ×10²¹ atoms of Pt to moles of Pt, so we multiply by this value.

$4.99 \times 10^{21} \ atoms \ Pt *\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

Flip the ratio so the units of atoms of platinum cancel.

$4.99 \times 10^{21} \ atoms \ Pt *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23} \ atoms \ Pt}$

$4.99 \times 10^{21} *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23}}$

$\frac { 4.99 \times 10^{21}}{6.022 \times 10^{23}} \ mol \ Pt$

Divide.

$0.008286283627 \ mol \ Pt$

<h3>2. Convert Moles to Grams </h3>

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. This is found on the Periodic Table because it is equivalent to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up platinum's molar mass.

Pt: 195.08 g/mol

Set up another ratio using this new information (195.08 grams of Pt in 1 mole of Pt).

$\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

Multiply by the number of moles we just calculated.

$0.008286283627 \ mol \ Pt*\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

The units of moles of platinum cancel.

$0.008286283627*\frac {195.08 \ g \ Pt}{ 1 }$

$0.008286283627* {195.08 \ g \ Pt}$

$1.61648821\ g \ Pt$

<h3>3. Round</h3>

The original measurement of atoms ( 4.99 ×10²¹ ) has 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 6 in the thousandth place tells us to round the 1 up to a 2.

$1.62 \ g \ Pt$

There are approximately <u>1.62 grams of platinum</u> in 4.99 ×10²¹ atoms of platinum.

8 0

3 years ago

How much charged is required to reduced 2 moles of cr6+ to cr3+

BartSMP [9]

Answer:

6e⁻

Explanation:

Charge in reducing 2 mole of Cr⁶⁺ to Cr³⁺

Cr⁶⁺ + xe⁻ → Cr³⁺

The numbers 6+ and 3+ on the chromium atoms are the oxidation number of the atom.

It shows the number of electrons that has been lost by chromium in an oxidation state.

Cr⁶⁺ shows that it has lost 6 electrons. When an atom loses electrons, the number of protons becomes more. This makes it positively charged in nature.

This analogy goes for Cr³⁺

Therefore, in going from reactants to products, chromium

6+ → 3+

In this kind of expression, the number of atoms must be conserved and the charges too;

2Cr⁶⁺ + xe⁻ → 2Cr³⁺

given 2 moles of Cr;

To balance the charge;

2(6+) + x(-) = 2(3+)

12 - x = 6

x = 6

A charge of 6e⁻ is required to reduce 2 mole of Cr⁶⁺ to Cr³⁺

6 0

4 years ago

Which has the greater number of particles, 1 mole br2 or 80g br2?

KengaRu [80]

They have the same amount of particles.

If you convert 80g to moles, you will find that 80g of diatomic bromine is 1 mole.

3 0

4 years ago

What was the theoretical yield of a reaction if 50g are formed with a 10% yield?

Bess [88]

Answer: 500g

Explanation:

We are asked to find the theoretical yield of a reaction, and we are given the following information:

In order to find the the theoretical yield we must use the following formula:

We have to convert the percents to real numbers before the calculations. We can do it dividing the percent value into 100, so:

4 0

3 years ago