Find the pH and concentrations of (CH3)3N and (CH3)3NH+ in a 0.060 M solution of trimethylammonium chloride.?
Kb ?
Find the pH and concentrations of (CH3)3N and (CH3)3NH+ in a 0.060 M solution of trimethylammonium chloride? Need step by step help please?
pH = pKa - lg( (CH3)3NH+ / (CH3)3N)) need Ka
Find the pH and concentrations of (CH3)3N and (CH3)3NH+ in a 0.060 M solution of trimethylammonium chloride.?
Kb ?
How can I find the concentration of H+ ions in a solution given its pH?
The pH of a solution is equal to the base 10 logarithm of the H+ concentration, multiplied by -1. If you know the pH of a water solution, you can use this formula in reverse to find the antilogarithm and calculate the H+ concentration in that solution. Scientists use pH to measure how acidic or basic water is. A low pH value means water is acidic and a high value means it's basic, often referred to as alkaline. In acidic water, there is an increased concentration of positively charged hydrogen atoms, H+. This concentration determines the pH value.Enter into the calculator the pH value for which you intend to calculate the H+ concentration. For example, if the pH of your solution is 5, enter 5 into the calculator. pH values will almost always be between 0 and 14, so your number should be within this range.Multiply the value you just entered by -1. This is the first step toward calculating the concentration of H+ in the solution, based on the equation pH = (-1) log [H+], where "log" is short for base 10 logarithm and the square brackets around H+ stand for "concentration." Multiplying the pH by -1 puts this equation in the form log[H+] = - pH. In the example, you would multiply 5 by -1 to get -5.Take the base 10 antilogarithm (or "anti-log") of the value you just calculated. You can take the anti-log by using the 10^x key on the calculator. By doing this, you are changing the pH equation into the form anti-log (log[H+]) = anti-log (- pH). The two reverse operations (anti-log and log) on the left-hand side cancel each other out, leaving [H+] = anti-log (- pH). So the value you calculate in this step is the concentration of H+ in the solution. The units of this concentration are molarity, or moles H+ per liter of solution. The example with a pH of 5 would therefore have an H+ concentration equal to anti-log (-5) which equals 0.00001 moles/liter. (properties of anti-logs from ref 3)TipSome scientists prefer to use the formula H3O+ instead of H+, to show that the positive hydrogen atom typically combines with a neutral water molecule (H2O) to form H3O+, known as the hydronium ion.
Find the pH and concentrations of (CH3)3N and (CH3)3NH+ in a 0.060 M solution of trimethylammonium chloride? Need step by step help please?
pH = pKa - lg( (CH3)3NH+ / (CH3)3N)) need Ka
Chemistry help - Calculating the pH of a solution?
Question a) You have produced a propionic acid. potassium propionate buffer soloution. The pH of a buffer solution is determined using the Henderson - Hasselbalch equation: You need: pKa propionic acid = 4.87 Molarity of CH3CH2COOH = 0.085M Molarity of CH3CH2COOK = 0.06M Equation: pH = pKa + log ([salt]/[acid] ) pH = 4.87 + log (0.06/0.085) pH = 4.87 + log 0.706 pH = 4.87 + (-0.15) pH = 4.72 Question b) you have a basic buffer. Easier to solve for pOH and convert to pH The Kb value for (CH3)3N is 6.5 x 10-5. pKb = -log (6.5*10^-5) = 4.19 pOH = pKa + log (salt]/[base]) pOH = 4.19 + log ( 0.10/0.075) pOH = 4.19 + log 1.33 pOH = 4.19 + 0.12 pOH = 4.31 pH = 14.00- pOH pH = 9.69 Question c: Once again you hace an acetic acid/ sodium acetate buffer pKa for CH3COOH = 4.74 Molarity of Acetic acid - the final volume is 100mL M1V1 = M2V2 M1*100 = 0.15* 50 M1 = 0.15*50/100 = 0.075M Molarity of CH3COONa M1*100 = 0.20/50 = 0.10M Now substitute into the H-H equation: pH = pKa + log ([salt]/[acid]) pH = 4.74 + log( 0.10/0.075) pH = 4.74 + log 1.33 pH = 4.74 + 0.12 pH = 4.86 You must be given the pKa or pKb values in an exam, or a reference list that allows you to access this data. You should not be required to remember these values.
How can I find the pH of a solution of NaHCO3?
pH of Amphiprotic species:NaHCO3 is amphiprotic species just like NaHSO4, NaHS, Na2HPO4, NaH2PO4. etc.In such cases the solution pH is dependent on the two ionization constants of the amphiprotic substance and independent of the concentration of the amphiprotic substance in the solution. However, they do not hold pH well and are not effective as buffer solutions. Here the trick is to use the formula,pH = (pKa1 + pKa2)/2. where Ka1 & Ka2 are dissociation constants of the conjugate acid of amphiprotic species.For NaHCO3, pH is (6.37 + 10.25)/2 = 8.31As considered for salts of H3PO4 (pKa1=2.148, pKa2=7.21, pKa3=12.67)-pH of (H2PO4)- , pH= (pka1 + pka2)/2 hence, For NaH2PO4, pH = 4.67.For Na2HPO4, pH is 1/2(pKa2+pKa3) = (7.21 + 12.67)/2 = 9.77.Isoelectric pH of aminoacids:Aminoacids are also an example of amphiprotic species. When an amino acid such as glycine, H2NCH2COOH, is dissolved in water. the carboxylic acid group loses a proton which is gained by the more basic amine group. This produces an ionic structure with opposite charges on both ends, a zwitter-ion(dipolar ion). The zwitter-ion structure of glycine is +H3NCH2COO-. The protonated form of this amphiprotic zwitterion, +H3NCH2COOH, is the glycinium ion. These amino acids are characterised by two pKas, pKa1 and pKa2 for the carboxylic acid and the amine respectively.Isoelectric pH (pI) is the pH at which aminoacids exists as zwitter-ion. The isoelectronic point(pI) will be halfway between, or the average of, these two pKas, i.e. pI = 1/2 (pKa1 + pKa2). For the simplest amino acid, glycine, pKa1= 2.34 and pKa2 = 9.6, hence pI =(2.34 + 9.6)/2 = 5.97.Important- (a) Fifteen of 20 amino-acids have isoelectric point near 6. (b) At isoelectric pH, all the molecules have equal positive & negative charges. Hence their is no migration towards any electrode during electrolysis at isolectric pH.For deivation of pH of amiphiprotic species:Special type- Let us consider the hydrolysis of amphiprotic anion along with cation, e.g., NH4HCO3, NH4HS. In above examples both cations and anions are derived from weak base and weak acids respectively hence, both will undergo hydrolysis in aqueous medium. When these salts are dissolved in water, [H3O+] concentration can be determined as,[H3O+] = √ka1[kw/kb + ka2]pH = -log = √ka1[kw/kb + ka2]Example-Reference: pH of Amphiprotic species & Isoelectric pH
How do I calculate the hydrogen ion concentration given the pH?
I hate writing long answers like others do for upvotes.*It’s Hydromium ion not hydrogen ion. Right? pH formula is always used for hydromium ion.Well I assume you know what is pH. If not then you can ask me. the formula ispH= -log [H3O+]if you are given the pH then you do the log inverse on both sides of the equation. So if you are given a pH of 3, then you will do the following:-3= -log[H3O+]When you multiply both sided by -1, then you will get,-3= log[H3O+]log inverse is to raise -3 to the base 10.10^-3 = [H3O+]0.001 = [H3O+]So in this case pH of hydromium ion is 0.001 MHope this helps!
Calculate the pH of a 0.10 M CH3NH3Cl solution?
CH3NH3+ is the conjugate acid of methylamine, CH3NH2. Kb for CH3NH2 = 4.4 x 10^-4. we need to calculate the Ka for CH3NH3+. Ka x Kb = Kw Ka = Kw / Kb = (1 x 10^-14) / (4.4 x 10^-4) = 2.3 x 10^-11 So when you put the acidic salt CH3NH3Cl in water, it will hydrolyze to form small amounts of H3O+ and CH3NH2. Molarity . . . .CH3NH3+ + H2O <==> H3O+ + CH3NH2 initial . . . . . . . .0.10 . . . . . . . . . . . . . 0 . . . . . . .0 change . . . . . . .-x . . . . . . . . . . . . . . x . . . . . . .x at equilibrium .0.10 - x . . . . . . . . . . . .x . . . . . . .x Ka = [H3O+][CH3NH2] / [CH3NH3+] = x^2 / (0.1 - x) = 2.3 x 10^-11 Because Ka is so small, x will be negligible compared to 0.10 so we can drop the x from the 0.10 - x term. x^2 / 0.1 = 2.3 x 10^-11 x^2 = 2.3 x 10^-12 x = 1.5 x 10^-6 = [H3O+] pH = -log [H3O+] = -log (1.5 x 10^-6) = 5.82
Calculate the pH of a 0.800 M CH3NH3Cl solution?
CH3NH3+(aq) <-------> CH3NH2(aq) + H+(aq) Ka = [H+]*[CH3NH2]/[CH3NH3+] if x moles of CH3NH3+ dissociate, then [H+] = x [CH3NH2] = x [CH3NH3+] = 0.800 - x Ka = x²/(0.800 - x) ≈ x²/0.800 Ka = 10^-14/Kb = 2.7*10^-11 x = [H+] = √[0.800*2.7*10^-11] = 4.65*10^-6 pH = 5.33