Hello, assalamualaikum warahmatullahi wabarakatuh, back again on the smart chemistry channel, friends , this time we will discuss material about buffer solutions. The buffer solution is also called a buffer solution or it may also be called a buffer solution. What is a buffer solution? so friends the buffer solution is a solution that can maintain pH. For example, I have a buffer solution with a pH of 4, so when I added a buffer solution whose pH was 4, I added acid or slightly wet or with water, the pH of the buffer solution, which was 4, did not change or tended to remain friends with the buffer solution. there are two types, there is an acid buffer solution and there is also an alkaline buffer solution, an acid buffer solution. It contains a mixture of fatty acids and their salts or their conjugate base. For example, what is an example of a weak acid, for example CH3COOH or acetic acid, if it becomes a buffer solution, this acetic acid solution must be mixed right with the salt means the example of the salt from CH3COOH is CH3COONa or it may also be between formic acid HCOOH and its salt salt from formic acid Means it can be HCOONa, HCOOK can also be yes then the second type of buffer solution is a basic buffer solution the base buffer solution consists of a mixture of weak bases and its salt or conjugate acid for example is b weak acid, for example, here NH3OH if it becomes a buffer, it means that it must be mixed with the salt. The salt means that NH4Cl can be like that, friends, or the salt is also called the conjugate acid of a weak base. We can make the buffer solution in two ways, namely by making it directly. directly from what was a mixture of a weak acid and its salt or mixing a weak base and its salt directly, now besides the direct method there is also what is called an indirect way so the buffer solution can also be made indirectly by reacting excess weak acid with a base strong so it's an excess of weak acid, this means that the number of moles of the number of moles of the weak acid is greater than the number of moles of the strong base. For example, like this, let's make a buffer solution by reacting 1 mole of CH3COOH with half a mole of Koh. CH3COOH plus this KOH he will produces CH3COOK + H2O where the CH3COOH was first one mole, not only half a mole, this one doesn't exist yet now this is the first one that reacts, of course, what reacts is the smaller one, which means that the reactant is half a mole because this coefficient is one all yes, the meaning is the same, it means that it is on the right, which is half, then we calculate the remainder . you can see that this clean is half a mole of CH3COOH and half a mole of CH3COOK this is a mixture of an acid buffer solution Yes because of the weak acid earlier then the indirect way to make a base buffer solution is by reacting a weak base especially with a strong acid so the excess, the larger model is the good one, the weak example is when we react with 1 mole of NH4OH Now with half a mole of HCl means that there is NH4OH plus HCl to become NH4Cl and H2O. NH$OH only one mole of this is first then reacts then the rest how much is this HCl half a mole at first NH4Cl and H2O no one has reacted yet how much reacts, of course the smaller one is half a mole, yes, friends, this is half a mole then HCl, yes, half a mole of NH4Cl because the coefficients are both one, which means this is also half a mole, then this is also half a mole of H2O, which means we can calculate the remaining one minus half , meaning this is half a mole, then this is finished, then the NH4Cl is 10 plus half. if the H2O is also half a mole Now from here we see that what is left is NH4OH, which is half a mole and NH4Cl, which is 0.5 mol NH4OH is a weak base . made from a weak base , we will go to the problem so that it is easier to understand the example of the first problem, a buffer solution can be made by mixing the dotted solution . Weak and its salt means from options a to e, friends, look for a weak acid or a weak base and the salt, we go to option a. hno3 and ch3coona ,HNO3 this is a strong acid meaning it is not b.hno3 and nano3 obviously hno3 is a strong acid yes that means it's not then that CH3po4 and CH3COONa he is indeed a weak acid but ch3coona is a salt but not of H3Po4 meaning this is not a pair which is a buffer solution then D NH4OH is a weak base then we check the salt it turns out to be ch3coona yes this is indeed salt but not salt that comes from NH4OH means it's not a partner so this is also not then we go to option e. NH4OH and NH4Cl. NH4OH this is a weak base nh4Cl is a salt of nh4oh so this is the correct answer so the answer is e. 2. Which of the following mixtures will form a buffer solution? Now this means that the indirect method, friends, we remember that for the indirect method, it can be done in two ways, namely by reacting a weak acid and a strong base with this weak acid in excess or it can also be done by reacting a weak language and a strong acid. what you want is a weak language, now this means that plants just use these instructions to answer question number 2 OK, let's check it from option A100 milli n NaOH + 100 milli HCL NaOH this is a strong base HCL this is a strong acid means it's impossible because this is not compatible with earlier, friends, then the B 50 ml NaOH 0.1 molar plus fifty milli CH3COOH 0.2 molar NaOH this is a strong base CH3COOH is a weak origin, now this is in accordance with this one, a weak acid, a strong base, but even though it's according to a friend -friends, you have to calculate the motor first, is it true that the moles of the weak acid are in excess. Now, the way to calculate the moles is by flowing the volume and concentration, so to count ung mole = M3 livic, let's calculate malenna Oh, first the v is fifty milli m, 0.1 means the mall is perfect, just multiplied by 50 times 0.1 = 5 millimoles, Enno is only five, always at least the mole of CH3COOH is fifty milli times 0.2, it means the model is 50 millimoles, 0.2 means the answer is 10 millimoles of weak acid, namely CH3COOH is 10 millimoles, while the moles of strong base are 5 millimoles, more than those of weak acids, it means that it is in accordance with this, meaning the answer is B. checked C, D yes, maybe C is fifty milli H2so4 0.1 molar H2so4 it's because it's a strong acid yes it's a strong acid then fifty milligrams of na2so4 na2so4 this is salt, salt Was it in this earlier no Yes It means this is wrong then the d100 milli NH3 0.1 molar plus 100 milli HCl 0.1 molar NH3 this is a weak base HCl this strong acid goes up following the second one Yep BL + AK weak base + strong acid but remember we have to also count the number of moles it turns out that the moles are the same friends y NH3 100 times 0.1 means 10 millimoles, HCl 100 times 0.1 is also 10 millimoles means it runs out later, if it runs out it's not a buffer solution but hydrolysis