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IgorC [24]
3 years ago
13

A solution of 0.2 M Sulfuric acid is titrated with a 0.2 M basic solution. In three to five sentences, explain how you can use t

he results of the titration as evidence for the ratio of sulfuric acid and the base in the balanced chemical reaction for the titration.
Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
5 0

If there are 0.2 M solutions of both acid and base, the concentrations of hydrogen and hydroxide ions will be equal at equivalence point.

The reaction of sulfuric acid and a basic solution BOH occurs as follows;

H2SO4(aq) + 2BOH(aq) -----> B2SO4(aq) + 2H2O(l)

In the question, we are told that that both the solution of the sulfuric acid and the basic solution are 0.2 M.

The point where all the hydrogen and hydroxide ions have reacted according to the stoichiometry of the reaction. If there is really equimolar amounts of acid and base, the concentration of hydrogen and hydroxide ions will be equal at equivalence point.

Learn more: brainly.com/question/2192784

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The rate of disappearance of hbr in the gas phase reaction 2hbr(g)?h2(g)+br2(g) is 0.160 m s-1 at 150°c. the rate of appearance
luda_lava [24]
The reaction rates of the substances whether disappearance of a reactant or the appearance of a product are related to each other by the chemical reaction. The reaction rates are related as follows:

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3 years ago
A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

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