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Serga [27]
3 years ago
6

Popeye wants to make a dilute spinach solution for Sweetpea's bottle. How much 3.0 M spinach solution should he add to the 500.0

mL bottle if he wants the final solution to be 0.30 M?
Chemistry
1 answer:
Lynna [10]3 years ago
8 0

Answer : The volume of 3.0 M spinach solution added should be, 50 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of spinach solution.

M_2\text{ and }V_2 are the final molarity and volume of diluted spinach solution.

We are given:

M_1=3.0M\\V_1=?\\M_2=0.30M\\V_2=500.0mL

Now put all the given values in above equation, we get:

3.0M\times V_1=0.30M\times 500.0mL\\\\V_1=50mL

Hence, the volume of 3.0 M spinach solution added should be, 50 mL

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A solution is prepared by dissolving 0.23 mol of chloroacetic acid and 0.27 mol of sodium chloroacetate in water sufficient to y
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Answer:

c. chloroacetate ion

Explanation:

The chloroacetic acid, ClCH₂CO₂H, is a weak acid with Ka = 1.36x10⁻³. When this weak acid is in solution with its conjugate base, ClCH₂CO₂⁻ (From sodium chloroacetate) a buffer is produced. The addition of a strong acid as the HCl produce the following reaction

HCl + ClCH₂CO₂⁻ → ClCH₂CO₂H + Cl⁻.

Where the acid reacts with the chloroacetate ion to produce more chloroacetic acid

That means, the HCl reacts with the chloroacetate ion present in the buffer solution

Right answer is:

<h3>c. chloroacetate ion</h3>
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A geologist finds a mineral sample alongside a swift flowing river. She determines the mineral is a type of Quartz. Which is tru
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Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the
kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH  = 3.40+ log {0.170 /0.105}

pH =  3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol  = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

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2 years ago
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