Convert the following 4. 100 cm = 5. 453 m = _ 6. 89847nm =

A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of

Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0

3 years ago

500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C

dangina [55]

Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v= $\frac{1,000 g CaCl2}{500mL}$ ×100 = 0,2%w/v

b) Ratio strength is a way to express concentration. For w/v is in 1g of solute r mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × $\frac{1mol}{110g}$ = 9,09×10⁻³ moles of CaCl₂

500 mL of solution × $\frac{1L}{1000mL}$ = 0,500 L of solution

M = $\frac{9,09x10^{-3} moles }{0,500 L}$ = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × $\frac{0,8 kg}{1L}$ = 0,625 kg of solution

m = $\frac{9,09x10^{-3}moles }{0,625 kg}$ = 0,0145 m

e) In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L × $\frac{0,0182 CaCl2 moles}{1L}$ × $\frac{1equivalent}{2 moles}$ = 0,0137 equivalents