Calculate The PH Of Acetic Acid For A 0.02 M Solution

Includes the "Assumption" teachers tell you to make. How do you find the pH of a weak acid solution? This video goes through the whole . Includes the "Assumption" teachers tell you to make. How do you find the pH of a weak acid solution? This video goes through the whole calculation. Just fKa(CH3CO2H) = 1.8 x 10-5. | Socratic. Course menu. Socratic Q&A logo. Search icon . The pH is roughly 8.96. Sodium acetate is the salt of a weak acid and strong base from the equation: C_(2)H_(3)NaO_2->CH_3COO^(-)+Na^(+), where: CH_3COO^(-)+H_2O\\rightleftharpoonsCH_3COOH+OH^(-) As it is a weak acid and strong base, this is a good indicator of a fairly high pH. K_b=([HB^+][OH^-])/([B]) where: [B] is the concentration of the base [HB^+] is the concentration of base ions. [OH^-] is the concentration of the hydroxide ions. Also, K_aK_b=1*10^(-14)/ So, K_b=(1*10^(-14))/(1.8*10^(-5))=5.555*10^(-10) [("",CH_3COO^(-),CH_3COOH,OH^-),(I,0.150,0,0),(C,-x,+x,+x),(E,0.150-x,x,x)] K_b=5.555*10^(-10)=x^2/(0.150-x) Since the value for K_b is small, we will assume the the value for x is small, and so will take 0.150. So, x=sqrt(0.150(5.555*10^(-10)))=OH^- pH=14-pOH=14-(-log(OH^-)) =14-(-log(sqrt(0.150(5.555*10^(-10))))) =14-(-log(9.128709292*10^(-6))) =14-5.039590623~~8.962.9×10−2M. Explanation: Let's start by writing the chemical reaction for the dissociation of silver phosphate:. 2.9xx10^(-2) M Let's start by writing the chemical reaction for the dissociation of silver phosphate: Ag_3PO_4(s) rightleftharpoons 3Ag^(+)(aq) +PO_4^(-3)(aq) Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3. Ksp=1.8xx10^(-5) M = [Ag^(+)]^(3) (aq) +[PO_4^(3-)] (aq) Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power. 1.8xx10^(-5) M = (3X)^(3) xx (X) (always multiply when finding the molar solubility). Now take 3^(3) which is 27 and divide the Ksp by 27, so you can get all of the X's by themselves. (1.8xx10^(-5) M)/27 = 6.67xx10^(-7)M Now you're left with X^(3) xx X, so multiply the X's to get X^(4). Take the fourth root of the 6.67xx10^(-7)M to obtain the value of x. (6.67xx10^(-7)M)^(1/4) = 2.9xx10^(-2) M The value of x that we just obtained is

What Is The PH Of A 0.150 M Solution Of Sodium Acetate (NaO2CCH3

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. What is the pH of a 0.205 M ammonia solution? Part B: What is the percent we're going to compare ammonia to annalen so nh3 is ammonia and the KB for ammonia is 1.8 times 10 to the negative 5 and aniline is c6h5 NH 2 and notice . Quick overview of Kb and pKb. Examples of calculating the pH of a weak base solution. . Quick overview of Kb and pKb. Examples of calculating the pH of a weak base solution.pH=11.13. Explanation: In aqueous solution, NH3 reacts with water according to the following reaction: NH3(aq)+H2O(l)→NH+4(aq)+OH−(aq). color(blue)(pH=11.13) In aqueous solution, NH_3 reacts with water according to the following reaction: " " " " " " " "NH_3(aq)+H_2O(l)->NH_4^+(aq)+OH^(-)(aq) Initial" " " "0.10M" " " " " " " " " " " "0M" " " " " "0M "Change" " " " "-x" " " " " " " " " " " "+x" " " " " "+x Equilibrium" "(0.10-x)M " " " " " " ""+x" " " " " "+x The equilibrium constant is written as: K_b=([NH_4^+][OH^(-)])/[NH_3]=1.8xx10^(-5) Replacing the equilibrium concentrations by their values in the expression of K_b: K_b=((x)(x))/((0.10-x))=1.8xx10^(-5) since the value of K_b value is small, we consider x"<<"0.10 Solving for x, x=1.34xx10^(-3)M x represents the concentration of OH^-. Using the expression of K_w=[H^+][OH^-] =1.0xx10^(-14) [H^+]=(K_w)/([OH^-])=(1.0xx10^(-14))/(1.34xx10^(-3))=7.46xx10^(-12)M Then, the pH is calculated by: pH=-log([H^+]) => pH=-log(7.46xx10^(-12)) color(blue)(pH=11.13)

At 25 °C, The Molar Solubility Of Silver Phosphate Is 1.8 × 10^-5 Mol L

For acetic acid, HC2H3O2, the Ka value is 1.8×10−5. Calculate the concentration of H3O⁺ in a 0.3 M solution of HC2H3O2. Solution. \(K_a\) is an acid dissociation constant, also known as the acid ionization constant. It describes the likelihood of the compounds and the ions to break apart from each other. As we …(c) Na2CO3. Explanation: (a). This is a weak acid: HOAc⇌H++OAc−. This is a long question so I will go straight to this expression for the pH of . (c) sf(Na_2CO_3) (a) This is a weak acid: sf(HOAcrightleftharpoonsH^++OAc^-) This is a long question so I will go straight to this expression for the pH of a weak acid, which can be derived from the ICE table: sf(pH=1/2[pK_a-loga]) sf(a) is the concentration of the acid. sf(pK_a=-logK_a=-log[1.8xx10^(-5)]=4.745) :.sf(pH=1/2[4.75-(-2)]) sf(pH=color(red)(3.37)) (b) This is the salt of a weak acid and a strong base so undergoes hydrolysis: sf(OAc^(-)+H_2OrightleftharpoonsHOAc+OH^-) sf(pK_a+pK_b=pK_w=14) :.sf(pK_b=14-4.75=9.255) We can use the corresponding expression for the pOH of a weak base: sf(pOH=1/2[pK_b-logb]) sf(b) is the concentration of the base. sf(pK_a+pK_b=14) :.sf(pK_b=14-4.475=9.255) :.sf(pOH=1/2[9.255-(-2)]=5.627) sf(pH+pOH=14) :.sf(pH=14-5.627=color(red)8.37) (c) Carbonate ions are weak bases: sf(CO_3^(2-)+H_2OrightleftharpoonsHCO_3^(-)+OH^-) sf(K_(b1)xxK_(a2)=K_w=10^(-14)) sf(K_(b1)=10^(-14)/(5.6xx10^(-11))=1.786xx10^(-4)) sf(pK_(b1)=-log[K_(b1)]=-log[1.786xx10^(-4)]=4.748) sf(pOH=1/2[pK_(b1)-lNote that the inputs are standard notation numbers. The answers are formatted in scientific notation and E notation. 122500 + 3655 = 1.26155 x 10. Scientific . Scientific notation calculator to add, subtract, multiply and divide numbers in scientific notation. Answers are provided in scientific notation and E notation/exponential notation. Basic Calculator Calculators > Math > Scientific Notation Calculator Scientific Notation Calculator Scientific Notation Calculator enter numbers or scientific notation Operand 1 Operator +We have a solution CHX3COOH (acetic acid) with c=0.02 mol/L and Ka(CHX3COOH)=1.8⋅10−5. Calculate the pH of this solution. All I know is that

Hi, Elle.

The Kb is the equilibrium consistent for the reaction of the base ammonia combining with water to supply ammonium, the conjugate acid, and a hydroxide anion (OH-).

Here is the response: NH3 + H2O --> NH4+ + OH-

The components for any equilibrium consistent is Okay= (product of goods)/ (manufactured from reactants) with gadgets of concentration in molarity or pressure (for gasses only). Pure elements like H2O don't seem to be included in the formulation.

The pH is immediately similar (as we'll see later) to the focus of OH- which is likely one of the merchandise.

Ok= (product of goods)/ (manufactured from reactants)

1.8×10^−5 = (OH-)(NH4+) / (NH3)

The given 0.205 M ammonia is the whole concentration for each types of that species, the acid and the bottom, NH3 and NH4+. However, for a weak base like ammonia, we can assume that the amount of NH3 that has been changed to NH4+ is so low (by way of a magnitude 5x less than 10, given via the kb) that doing the calculation would not alternate the solution through very much in any respect. Going back to the chemical equation, we will be able to see that for every 1 mol of NH3, 1 mol of NH4+ and 1 mol of OH- is made. In different phrases, the amount of NH4 and OH- will always be the similar as each and every different here, so we will be able to use one variable "x" to constitute them each.

1.8×10^−5 = (OH-)(NH4+) / (NH3)

1.8×10^−5 = (x)(x) / (0.205 M)

1.8×10^−5 = x^2 / (0.205 M)

Solving for x, x = 0.00192 M where x = the focus of OH-

To get from the concentration of the OH- to the pH we want to know that the "p" merely way "the negative log of the concentration" and that pOH + pH = 14 all the time.

pOH = the adverse log of the focus of the OH = - log (0.00192 M) = 2.72

pOH + pH = 14 so pH = 11.3

For phase A, Final solution: pH = 11.3

The next part of the query asks for the percent ionization. Notice within the chemical equation that NH3 is going from having no price to generating two charged species. The p.c ionization is in point of fact simply asking what % of the NH3 becomes NH4+ and OH-. We already solved for the focus of the OH- which was once 0.00192 M. So 0.00192 M of the unique 0.205 M used to be ionized. 0.00192 is 0.937 p.c of 0.205 M.

For section B, Final resolution: 0.937%