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lana66690 [7]
3 years ago
14

Titration of 20.0 mL 0.120 M HCHO2 with 0.0800 M NaOH. Find the pH: (a) before NaOH addition; (b) after addition of 15.0 mL NaOH

; (c) after addition of 30.0 mL NaOH; (d) after addition of 30.5 mL NaOH.
Chemistry
1 answer:
stich3 [128]3 years ago
5 0

Explanation:

Given:

Ka = 1.8 × 10-4

HCOOH --> HCOO- + H+

HA --> H+ + A-

Ka = [H+] × [A-]/[HA]

[HA] = 0.12 M

[H+] = [A-] = x^2

1.8 × 10^-4 × 0.12 = x^2

x = 0.00465 M

pH = - log [H+]

= -log [0.00465]

= 2.33

B.

Since 1 mole of NaOH reactes with 1 mole of HCOOH.

Number of moles of NaOH = 0.08 × 15 × 10^-3

= 0.0012 mole.

Concentration of HCOOH = 0.0012/20 × 10^-3

= 0.06 M

[HA] = 0.06 M

1.8 × 10^-4 × 0.06 = x^2

x = 0.00329 M

pH = - log [H+]

= -log [0.00329]

= 2.48

C.

Since 1 mole of NaOH reacts with 1 mole of HCOOH.

Number of moles of NaOH = 0.08 × 30 × 10^-3

= 0.0024 mole.

Concentration of HCOOH = 0.0024/20 × 10^-3

= 0.12 M

[HA] = 0.12 M

1.8 × 10^-4 × 0.12 = x^2

x = 0.00465 M

pH = - log [H+]

= -log [0.00465]

= 2.33

D.

Since 1 mole of NaOH reactes with 1 mole of HCOOH.

Number of moles of NaOH = 0.08 × 30.5 × 10^-3

= 0.00244 mole.

Concentration of HCOOH = 0.00244/20 × 10^-3

= 0.122 M

[HA] = 0.122 M

1.8 × 10^-4 × 0.122 = x^2

x = 0.00469 M

pH = - log [H+]

= -log [0.00469]

= 2.33

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A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?
kotykmax [81]
First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)

we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters

we can now isolate mass and get
M * Liters * molar mass = mass

now we need to find the molar mass of CaCl2 which is 110.98 g/mol

plug the values in and get
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5 0
3 years ago
Read 2 more answers
Using the van der waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 c is____________at
MatroZZZ [7]

Answer:

D. 1.48atm

Explanation:

Van der waals equation is given as:

(P +an²/v²) (v - nb) = nRT

Where;

P = pressure (atm)

V = volume (L)

R = gas constant (0.0821 Latm/molK)

a and b = gas constant specific to each gas

T = temperature (K)

n = number of moles

According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol

Hence;

(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15

(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638

(P + 0.0291) (22.316) = 33.638

22.316P + 0.649 = 33.638

22.316P = 33.638 - 0.649

22.316P = 32.989

P = 32.989/22.316

P = 1.478

P = 1.48atm

6 0
3 years ago
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