How many mL of 0.10 M NaOH should the student add to 20 mL 0.10 M HFor if she wished to prepare

How many mL of 0.10 M NaOH should the student add to 20 mL...

Key Concepts

Molarity, Volume, and Moles Relationship

The relationship between molarity, volume, and moles is fundamental in solution chemistry. Molarity (mol/L) multiplied by volume (L) gives the number of moles of a substance present. This concept is essential when calculating the amount of titrant (in this case, a strong base) needed to react with a given quantity of acid to achieve the desired buffer composition.

Acid-Base Neutralization

Acid-base neutralization refers to the reaction in which an acid reacts with a base to produce water and a salt. In the context of buffer preparation, this process is used to convert a portion of the weak acid into its conjugate base by adding a strong base, thereby creating the necessary components for a buffer system.

Buffer Solutions

A buffer solution is a mixture that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) and is capable of resisting changes in pH when small amounts of an acid or a base are added. Understanding buffer solutions is crucial for preparing a solution with a predetermined pH because buffers maintain stability by neutralizing added acids or bases.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a relationship between the pH of a buffer solution, the pKa of the weak acid, and the ratio of the concentrations of the conjugate base to the weak acid. This equation is central to calculating the necessary proportions of the components in a buffer to achieve a specific pH.

Transcript

00:01 Hi everyone so in order to tackle this problem you trade first given data now what they were given you know they provided k a for hf that is equal to 1 .8 into 10 dash to minus 4 secondly they provided ph ph is equal to 3 .55 then volume of an average they ask the next mularity of an hour .10 molar.

00:40 Next, volume of hf, 20ml.

00:48 The next, molarity of hf, 0 .10 mole.

01:00 Now, once you know this, then gone calculating a part, that is, pk, is calculation of pk.

01:15 Now what is pk you know pk is equal to negative log of concentration of k to the best 10 and that is equal to negative log of 1 .8 into 10 dash minus 4 so this is equal to amaz minus into minus 4 log of 10 it will become plus 4 minus love of 1 .8 and that is equal to 4 minus 0 .253 and that is equal to how much 3 .745 this is the value of pk now once you know the pk value value then calculate volume of an image using volume of an image you have to calculate using in henderson hesel belize equation and there is a hesl bulge equation.

02:36 Now let us consider v be the volume of v be the volume of 0 .1 molar and hence.

02:55 Therefore total volume of solution solution that is equal to 20 plus v then you know concentration of hf that is equal to what 20 into point 1 divided by 20 plus x molar similarly concentration of salt that is equal to what 20 mm into point one divided by 20 plus v here it is not smaller then you know secondly henrys and heselbellis equation is given as ph is equal to pka plus log of concentration of a salt divided by concentration of acid to the best 10 so by studying the values in this equation what will get in a ph they provided 3 .55 that is equal to 3 .745 plus log of concentration of salt that is v into 0 .1 divided by 20 plus v divided by 20 into 0 .1 divided by 20 plus v to the best 10.

04:46 Therefore this is equal to what 3 .55 minus 3 .745 that is equal to log of 20 log of v .1 divided by 20 plus v into 20 plus v upon 20 into 0 .1.

05:18 Now, you know this 20 plus v, 20 plus v get cancelled...

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