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What is the concentration of free Agt (un-complexed Agt) in a 4.170 M solution of Ag(NH3hzt Kf of Ag(NH3h2' 1.7x 107 The reaction for the formation of Ag(NH3)z...

Question

What is the concentration of free Agt (un-complexed Agt) in a 4.170 M solution of Ag(NH3hzt Kf of Ag(NH3h2' 1.7x 107 The reaction for the formation of Ag(NH3)z is: Agt (aq) + 2 NH3 (aq) Ag(NH3)2

What is the concentration of free Agt (un-complexed Agt) in a 4.170 M solution of Ag(NH3hzt Kf of Ag(NH3h2' 1.7x 107 The reaction for the formation of Ag(NH3)z is: Agt (aq) + 2 NH3 (aq) Ag(NH3)2



Answers

A solution is made 1.1 * 10-3 M in Zn(NO3)2 and 0.150 M in
NH3. After the solution reaches equilibrium, what concentration
of Zn2+(aq) remains?

Problem. 16.53. We have to Ami and we want to find the KB or no not the KB. We want to find a hydroxide ion concentration given that the initial concentration of basis 0.60 So I set up an ice box here. You subtract X from the reactant and add X to the products to get the equilibrium concentration. So this will be approximately 0.60 Because I can use the assumption that X is negligible since the K be developed or the initial concentration divided by the KB is greater than 100. So that's why you can use this assumption. So this will be X. And this will also be X. And you just put this into the KB formula. I looked up the KB from one of the tables and the chapters. So you just do 4.4 times 10 to the negative four equals X squared the concentration. And the products divided by the reactant since their 0.60 And when you saw for ex you get X equals 0.51 So that would equal the O H minus concentration right here. So the answer will be 0.51 And the second part of the question asked what is the ph so to find the ph you simply just you have to first find the p O. H since we haven't O H minus concentration. So the way to do that is you do people h mind equals the negative log of the h minus concentration which is 0.51 And this will give you the P O. H of around 2.29 And the ph would just be 14 minus the p or P O H 2.29 which will give you 11.71 And you want to think things here and ph you only look at those think things after the decimal point. So these two will be your answer right here.

We know that the case p for gold chloride from the balanced chemical reaction is concentration of au multiplied by the concentration of chloride because the reactions 1 to 1, the concentration of gold will be equal to the concentration of chloride. If you look up K SP, it's 2.0 times tend to negative 13. That then is going to be equal to the gold concentration multiplied by the chloride concentration. But the chloride concentration is the gold concentration, so it's just gold concentration squared. So the gold concentrations just the square root of Caspi or 4.5 times 10 the negative seven Moeller.

This question asked for the ph of a 0.35 molar solution of the method, ammonium chloride. We recognize that methyl ammonium is a an acid, being the conjugate acid to methyl amine. So to calculate the ph we can set up the equation the hydro Nehemiah concentration because this is an acid will be equal to the K. A. Value for the methyl ammonium ion, which is the KB value for methyl amine, which we can look up in the back of the book, divided into K. W. We then multiply the K value by the concentration of the weak acid, the methane ammonia may in and we get a hydro knee um concentration of 282 times 10 to the negative six moller. The ph then is simply the negative log of the hydro knee um concentration Giving us a ph of 555. Because every time we make a hydro Nehemiah on um ethel ammonium has turned into a methyl amine. Then the hydro nia mind concentration is the method. I mean concentration

16.54. We have the weak base and H 20 H. And 0.21 more concentration. And we want to find the concentration of the hydroxide ion. And to do that, you have to set up an ice box first you subtract X from the reactant and add extra products. We can approximate this to 0.21 because the K. Um the initial concentrations are point to one divided by the KB is going to be greater than 100 this will just be X. So you plug it into the KB expression which is the concentration of O H minus times the concentration of NH studio H plus over the concentration of the base and H. 20 H. So that would just be the KB. Which I looked up from the textbook 1.1 times 10 to the negative eight equals X squared over 0.21 And when you solve for X we get that X is 4.80 It one times 10 to the negative five Moeller. So that would be the concentration of O h minus and you need to 66 So it would just be 4.8 times 10 to the negative five Moeller. Mhm. So that would be your first answer and the second part as you to solve the ph of the solution. So to find the ph you first have to find the P. O. H. Since we have the hydroxide ion concentration, not the hydro sodium ion concentration. So to find the P. O. H, you just do the negative log of the hydroxide ion concentration which we found in a previous part, which gives you 4.31 as a P 32 as the people age. So the ph would just be 14 minus 4.32 which will give you 9.68 and you want to six figs because, and then ph and p o h. You look at the sick things after the decimal point, so these two would be your answer right here.


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