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3 years ago

PLS ANSWER ASAP, WILL GIVE BRAINLIEST ANSWER

Chemistry

1 answer:

Stolb23 [73]3 years ago

6 0

Answer:

.14L or 140mL

Explanation:

This is a classic plug-n-chug problem. Your textbook probably goes over this formula as $M_{1}V_{1} = M_{2}V_{2}$ . M stands for molarity of the given substance, and V stands for the volume that the substance occupies.

Simply plug in the values that you're given, like so:

$M_{1} = 1.5M\\V_{1} = ???\\M_2 = 7M\\V_2 = .03L$

$1.5M * x = 7M * .03L$

After completing the algebra portion and solving for the unknown, you will be left with x = .14L, which is the volume required to neutralize 30mL of 7M NaOH.

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3 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

<em> </em>inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$

Therefore, the new pH is 7.69.

Learn more about pH here:

brainly.com/question/24595796

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