How many moles can be found in 1.5 x 10^24 atoms of Calcium?

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

Which statement is most likely a scientific law?

Annette [7]

Answer:

we need to see the statements

Explanation:

6 0

3 years ago

Read 2 more answers

A biofuel researcher is running an experiment to produce methane from wastewater treatment plant sludge. The researcher places t

vodomira [7]

Answer: Missing data related to question is attached below

If the KH,CH4 = 1.42 * 10^-3 mol/L atm

answer:

0.88 atm

Explanation:

P1 = 20 psi

aqueous concentration of methane in closed bottle = 20 ppm

20 ppm = 20 / 1000 ( gram / Kg ) = 0.020 gram/kg = 0.020 g/liter

molar mass of methane = 16 gram/mol

<em>next : convert mass conc to molar conc </em>

Cm = 0.020 / 16

Cm = 1.25 * 10^-3 mol/L

Given that KH,CH4 = 1.42 * 10-3 mol/L atm

Applying equilibrium relationship

Cm = ( KH,CH4 ) ( partial pressure of methane )

hence partial pressure of methane

= Cm / ( KH,CH4 ) = 1.25 * 10^3 / (1.42 * 10^-3 ) = 0.88 atm

7 0

3 years ago

Phosphate buffer solution is purchased as a(n) 10x stock solution. The solution is more concentrated than the working stock. If

TiliK225 [7]

Answer:

v₁ = 7.5mL

Explanation:

using the formula

m₁v₁=m₂v₂

given m₁=10x m₂=1x v₁=? v₂=75mL

10x × v₁ = 1x × 75mL

v₁ = 1x × 75mL / 10x

v₁ = 7.5mL

So, 7.5mL of 10x PBS should be added with 67.5mL of water

6 0

4 years ago

Which process involved in deep water currents

kolbaska11 [484]

Its convection. i hope this helps please give me brainliest

4 0

3 years ago