Homework Solution: 1) Using the Henderson-Hasselbach equation, what will the pH become when buffers P1 and P2…

    ) Using the Henderson-Hasselbach equation, what will the pH become when buffers P1 and P2 are mixed during plasmid purification?____________ . Ignore the EDTA and the cells in your calculation and assume Tris-HCl has a pKa of 8.0. Information: Buffer P1: 50mM Tris-HCl, pH8, 10mM EDTA, .1mg/ml RNase A Buffer P2: Sodium hydroxide (200mM) and SDS (1%) Teachers tip: "the pKa of protonated Tris as 8.0. Since the NaOH concentration is larger than the Tris concentration the pH is based on the left over NaOH. Remember pOH+pH = 14." Using the information above and the Henderson-Hasselbach equation, what will the pH become when P1, P2, and N3 are mixed during plasmid purification? ___________ . Assume the pKa of acetic acid is 4.8 and remember to take into account the different volumes used. Information: Buffer N3: Acetate buffer (.9M, pH 4.8) to neutralize the sodium hydroxide, and guanidine hydrochloride (4.2M), which denatures protein. N3 provides a high salt concentration, so that the DNA sticks to the silica gel membrane in the small column. Volumes being used: P1: 250uL(microliter) P2: 250uL N3: 350uL There is a partial answer previously posted, but without any explanations to the procedure. Please explain the solution along with the work. Thanks!

    Expert Answer

    Answer

    1) Using the Henderson-Hasselbach equation, what conciliate the pH grace when buffers P1 and P2 are modified during plasmid disinfection?____________ . Ignore the EDTA and the cells in your reckoning and pretend Tris-HCl has a pKa of 8.0.

    Information: Buffer P1: 50mM Tris-HCl, pH8, 10mM EDTA, .1mg/ml RNase A

    Buffer P2: Sodium hydroxide (200mM) and SDS (1%)

    Teachers tip: “the pKa of protonated Tris as 8.0. Since the NaOH strain is larger than the Tris strain the pH is grounded on the left aggravate NaOH. Recall pOH+pH = 14.”

    Using the advice overhead and the Henderson-Hasselbach equation, what conciliate the pH grace when P1, P2, and N3 are modified during plasmid disinfection? ___________ . Pretend the pKa of acetic cutting is 4.8 and recall to admit into recital the divergent compasss used.

    Information: Buffer N3: Acetate buffer (.9M, pH 4.8) to counterpoise the sodium hydroxide, and guanidine hydrochloride (4.2M), which denatures protein. N3 provides a noble salt strain, so that the DNA sticks to the silica gel membrane in the slight column.

    Volumes nature used:

    P1: 250uL(microliter)

    P2: 250uL

    N3: 350uL

    There is a local apology previously posted, excluding extraneously any explanations to the proceeding. Please clear-up the explanation along with the production. Thanks!

    Expert Apology

    Answer

    1) The scrutiny is referable attributable attributable attributable very obviously presented; ultimately, from what I silent, you modified 250 µL of buffer P1 with 250 µL of sodium hydroxide (NaOH), which is referable attributable attributable attributable a buffer, excluding a very robust shameful. The buffer P1 consists of Tris-HCl and the conjugate shameful. The reaction initiative fix among Tris-HCl and NaOH is shown beneath.

    Tris-HCl (aq) + NaOH (aq) ——-> Tris (aq) + NaCl (aq) + H2O (l)

    NaCl and H2O do referable attributable attributable attributable join-in in the pH reckonings; hereafter permission them quenched.

    As per the balanced stoichiometric equation,

    1 controlbearance Tris-HCl = 1 controlbearance NaOH

    Forbearance Tris-HCl ascititious = (250 µL)*(1 L/106 L)*(50 mM)*(1 M/1000 mM)*(1 mol/L/1 M) = 1.25*10-5 mole.

    Forbearance NaOH ascititious = (250 µL)*(1 L/106 L)*(200 mM)*(1 M/1000 mM)*(1 mol/L/1 M) = 5.00*10-5 mole.

    Forbearance NaOH unreacted = (forbearance NaOH ascititious) – (forbearance Tris-HCl ascititious) = (5.00*10-5 mole) – (1.25*10-5 mole) = 3.75*10-5 mole.

    Entirety compass of the explanation = (250 + 250) µL = 500 µL.

    Molar strain of surplus NaOH in the explanation = (forbearance NaOH)/(entirety compass) = (3.75*10-5 mole)/[(500 µL)*(1 L/106 µL)] = 0.075 mol/L = 0.075 M.

    Since, we entertain surplus NaOH, we ascertain the pOH as pOH = -log [OH] = -log (0.075) = -(-1.1249) ≈ 1.125.

    Since pH + pOH = 14, we entertain pH = 14 – pOH = 14 – 1.125 = 12.875 (ans).

    Use the Henderson-Hasslebach equation to rate the harmony of the strains of acetic cutting/acetate in the buffer.

    pH = pKa + log [acetate]/[acetic cutting]

    ===> 4.8 = 4.8 + log [acetate]/[acetic cutting]

    ===> 0.0 = log [acetate]/[acetic cutting]

    ===> [acetate]/[acetic cutting] = 1 [log (1) = 0.0]

    ===> [acetate] = [acetic cutting] ……(1)

    Again, [acetate] + [acetic cutting] = 0.9 M

    ===> 2*[acetic cutting] = 0.9 M [from (1) overhead]

    ===> [acetic cutting] = [acetate] = ½*(0.9 M) = 0.45 M.

    The compass of acetate buffer ascititious is 350 µL; hereafter controlbearance acetate = controlbearance acetic cutting = (350 µL)*(1 L/106L)*(0.45 M)*(1 mol/L/1 M) = 1.575*10-4 mole.

    The acetic cutting in the buffer conciliate counterpoise the NaOH as per the reaction

    HAc (aq) + NaOH (aq) ——–> NaAc (aq) + H2O (l) [HAc = acetic cutting; NaAc = sodium acetate]

    As per the stoichiometry of the reaction,

    1 controlbearance HAc = 1 controlbearance NaOH = 1 controlbearance NaAc

    Therefore, 3.75*10-5 forbearance NaOH = 3.75*10-5 forbearance NaAc controlmed = 3.75*10-5 forbearance HAc counterpoised.

    Therefore, controlbearance of unreacted HAc at makeweight = (1.575*10-4 – 3.75*10-5) controlbearance = 1.2*10-4 mole; controlbearance of NaAc at makeweight = (1.575*10-4 + 3.75*10-5) controlbearance = 1.95*10-4 mole.

    Since the entirety compass of the explanation is the identical (500 + 350) µL = 850 µL control twain HAc and NaAc, hereafter, we can specific the molar strains of HAc and NaAc at makeweight as the appertaining estimate of controlbearances.

    Use the Henderson-Hasslebach equation intermittently.

    pH = pKa + log [acetate]/[acetic cutting] = 4.8 + log (1.95*10-4 mole)/(1.2*10-4 mole) = 4.8 + log (1.625) = 4.8 + 0.2108 = 5.0108 ≈ 5.011 (ans).