Is a octopus a carnivore or a herbivore

16.78 A buffer is prepared by mixing 525 mL of 0.50 M formic acid, HCHO2, and 475 mL of 0.50 M sodium for- mate, NaCHO2. Calcula

dolphi86 [110]

Answer: Before addition of HCl, the pH is 3.70 and after addition of HCl, the pH is 3.64 .

Explanation: The pH of the buffer solution is calculated by using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

Let's calculate the moles of acid and base(salt) present in the original buffer.

mL are converted to L and then multiplied by molarity to get the moles.

$525mL(\frac{1L}{1000mL})(\frac{0.50molHCHO_2}{1L})$

= $0.2625molHCHO_2$

$475mL(\frac{1L}{1000mL})(\frac{0.50molNaCHO_2}{1L})$

= $0.2375molNaCHO_2$

Total volume of the buffer solution = 0.525 L + 0.475 L = 1.00 L

Since, the total volume is 1.00 L, concentration of base will be 0.2375 M and the concentration of acid will be 0.2625 M.

pKa for formic acid is 3.74. Let's plug in the values in the equation and calculate the pH of the original buffer.

$pH=3.74+log(\frac{0.2375}{0.2625})$

pH = 3.74 - 0.04

pH = 3.70

Now, we add 8.6 mL of 0.15 M HCl acid to 85 mL of the buffer. Let's calculate the moles of acid and base in 85 mL of the buffer.

$85mL(\frac{1L}{1000mL})(\frac{0.2625molHCHO_2}{1L})$

= $0.0223molHCHO_2$

$85mL(\frac{1L}{1000mL})(\frac{0.2375molNaCHO_2}{1L})$

= $0.0202molNaCHO_2$

Now, let's calculate the moles of HCl added to the buffer.

$8.6mL(\frac{1L}{1000mL})(\frac{0.15molHCl}{1L})$

= 0.00129 mol HCl

This added HCl reacts with base(sodium formate) and formic acid is produced.

So, 0.00129 moles of HCl will react with 0.00129 moles of sodium formate to produce 0.00129 moles of formic acid. We can write formate ion in place of sodium formate and hydrogen ion in place of HCl. The equation would be:

$H^++CHO_2^-\rightarrow HCHO_2$

moles of base after reaction with HCl = 0.0202 mol - 0.00129 mol = 0.01891 mol

moles of acid after addition of HCl = 0.0223 mol + 0.00129 mol = 0.02359 mol

Let's plug in the values again in the Handerson equation and calculate the pH:

$pH=3.74+log(\frac{0.01891}{0.02359})$

pH = 3.74 - 0.096

pH = 3.64

8 0

3 years ago

what is the molarity of an HCI solution if 25.0 ml of 0.185 M NaOH is required to neutralize 0.0200 L of HCI?

butalik [34]

Answer:

Molarity of HCl solution = 0.25 M

Explanation:

Given data:

Volume of NaOH= V₁ = 25.0 mL (25/1000 = 0.025 L)

Molarity of NaOH solution=M₁ = 0.185 M

Volume of HCl solution = V₂ = 0.0200 L

Molarity of HCl solution = M₂= ?

Solution:

M₁V₁ = M₂V₂

0.185 M ×0.025 L = M₂ × 0.0200 L

M₂ = 0.185 M ×0.025 L / 0.0200 L

M₂ = 0.005M.L /0.0200 L

M₂ = 0.25 M

7 0

3 years ago

A weather balloon is filled with helium that occupies a volume of 5.00 × 10^4 L at 0.995 atm and 32.0°C. After it is released, i

DaniilM [7]

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

3 0

3 years ago

__ Cu + ___ HNO₃ → ___ _Cu(NO₃)₂ + ___ NO₂ + ___ H₂O

ivolga24 [154]

Answer:

Cu + 4HNO3 ---> Cu(NO3)2 + 2NO2 + 2H2O.

Explanation:

Balancing:

Cu + 4HNO3 ---> Cu(NO3)2 + 2 NO2 + 2H2O.

3 0

3 years ago

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Which of the following describes the relationship correctly?

galben [10]

A. Air expands, becomes less dense and rises when the temperature is high due to the increased kinetic energy. This causes the air molecules to move around much more so they are further apart.

8 0

3 years ago

Read 2 more answers