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4 years ago

A small amount of acid is added to a buffer solution, the ph of the solution will

Chemistry

2 answers:

Alik [6]4 years ago

7 0

Increase due to rising ph

Anastasy [175]4 years ago

3 0

A small amount of acid is added to a buffer solution, the ph of the solution will

A. increase

B. decrease

C. stay about the same

D. become neutral

correct option is <u>option C </u>

Buffer is a solution which can resist the change in pH with small addition of acid or base.

Hence a small amount of acid when added to a buffer solution will not change its pH,or its <u>pH remains the same.</u>

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A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

3 years ago

Someone help w one of these questions. Or if you know all of them surely help.

fenix001 [56]

Answer:

1 A true

B False

C not sure about it False

D False

5 0

3 years ago

What information must be specified when a chemist describes the solubility of a solute in water?

Serggg [28]

Answer:

Information that must be specified is the temperature of the solution,Amount of solute,Amount of solvent,and Identity of the solvent.

Explanation:

5 0

4 years ago

Measure and record the masses of all metal strips you set out in front of the test tubes

jeka94

Magnesium ribbon dissolves in zinc nitrate and copper nitrate solutions.

Even though the question is incomplete and refers to your practical work, however, I will try to help you as much as I can.

Metals dissolves in solutions of other metals that are lower than them in the electrochemical series. Hence, copper strip will show no change in magnesium nitrate or zinc nitrate solution. A zinc strip will not show any change in magnesium nitrate or zinc nitrate.

However, a magnesium ribbon will dissolve very quickly in zinc nitrate and copper nitrate solutions.

Learn more: brainly.com/question/14396802

3 0

3 years ago

The free rapid movement of electrons in metallic bonds makes them

babunello [35]

Answer:

C: Good conductors of electricity

Explanation:

By definition, Metals have metallic bonds and thus they are good conductors of electricity due to the fact that their electrons in the electron sea are usually free to flow and carry electric current. Thus, the correct answer is option C.

3 0

3 years ago