Question

A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach using gas chromatography mass spectrometry (GCMS). First attempts using external calibration proved difficult, as reproducibility and external standard agreement were poor. To overcome these problems, the researcher used the internal standard calibration method with chloroform as the internal standard. To begin, the researcher analyzed a sample containing 5.43 mg/L DDT standard and 3.07 mg/L chloroform as the internal standard, which produced peak areas of 4197 and 9861 for the DDT and chloroform, respectively. Determine the response factor (F). Calculate the pH and the pOH of an aqueous solution that is 0.050 Min HCl(aq) and 0.065 M in HBr(aq) at 25°C. Given that K for HCOOH is 1.8 x 104 at 25°C, what is the value of K, for COOH at 25 °C? A certain weak base has a K of 9.00 x 10-7. What concentration of this base will produce a pH of 10.23? Calculate the pH of 4.6 x 10- M HCl. Report your answer to the hundredths place. The measured pH of a 0.300 M HBr solution at 25°C is 0.632. Calculate the activity coefficient for H in this solution. Enough of a monoprotic weak acid is dissolved in water to produce a 0.0124 M solution. The pH of the resulting solution is 2.60. Calculate the K, for the acid.

Answer 1

- pH a por of 0.050 M HCl e 0,065 M HBr at 25°C [Hel] = [4+] = 0,05 M 2 Both are strong and [her] = l +] = 0.065 g (with lang ke Todel (H+) = 0.050 +0,065 - 0.1159 Now pH = -log T ht] = -log 0.115 CS Scanned with TpH-0.94 CamScanner

DLL and pola 14 - P44 - 14-0,94 poy = 13.06 З К Нcom | - ko for coot = | в хіб 1 kwa 10x10 - 1.88 104 So, sol 1 = s. 56 x 16" 5.56×101 & Ginen Ky of base 5 = 9.00 x167 pHa 10.23 pona 14 - 10.23 = 3.77 - 4 1.698810 - or - 10 since, - B tho M — ҡ M-1.6988154 BH + oH 0 0 tu I the с 1.6988107 1.695x10 Scanned with

kha [BH] Lon] (1.698810472 (M-1.698x104) 9.00 x 10' int 2.884 x 10 = M-16988154 9.00 x107 m - 1.528x100 = 2.884x108 -7. 9.00 X 18 M = 2.89928 X 100 M = 3.22 x 10 of weak so the base initial concentres bien in 10.0322 M Орн 1 + xit мне | Ką of Hel = 1.38108 Now, Helt the ci + thot tv 6.66108 - Y u с Scanned path 4.6 X16X

К. : сч) Lo+) (нс|] П - YO 4. Схіб - YO Nort 13xo 2 , (4-6 кг * ) 5. 98 x 16° — з я* x - - solning quadostic en ewe get х = 46 х 1 Сң, +] х= 4 с хв 1 pH - he'] -17. оооч. ) pH. 2.34

I have solved 4 questions for u...as I was running out of time ...pls repost rest of 2-3 questions...

And pls do upvote.