Bch367

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Arizona State University *

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Chemistry

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Feb 20, 2024

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Title: Buffer Preparations Results Weak acid, Strong acid, Weak base, and Strong base are combined to create buffer solutions with a pH of 7.00. A given pKa of phosphate buffer is 7.2. Potassium phosphate (K 2 HPO 4 174 g/mol) is the weak base, and potassium phosphate, monobasic (KH 2 PO 4 136 g/mol), as the weak acid. NaOH was the strong base. and HCl was the strong acid. Table 1 represents the buffer preparation calculation when acids and bases are paired in order to get the mass. Table 2 reviews results of the pH for each of the buffer types and their average pH. The ingredients were measured out with a target pH of 7.0, a volume of 50 mL, and a concentration of 0.1M using the Henderson-Hasselbach equation. Table 3-4 represents the calculated pH and the buffer capacity calculations. Table 1 Buffer preparation for process 1-3 Method 1 (WA+WB) Method 2 (WB +SA) Method 3 (WA+SB) Mass of KH2PO4 (g) 0.4216 g XXXXXXXXXXX 0.68g Mass of K2HPO4 (g) 0.330 g 0.871g XXXXXXXXXXXX Volume of 1 M HCI (mL) XXXXXXXXXXXX 3.1 mL XXXXXXXXXXXX Volume of 1 M NaOH (mL) XXXXXXXXXXXX XXXXXXXXXXXX 1.9mL Volume of H2O (mL) 50 mL 46.9 mL 48.1 mL

Final Volume (mL) 50 50 50 For Table 1 Henderson-Hasselbach equation was used for method 2 and 3. Stoichiometry allows the calculation of Method 1. The calculations were used with the pH given of 7.0, a volume of 50mL and 0.1M concentration. Calculations HA = H2PO4 = 136 g/mol A-= HPO4^2 = 174 g/mol 7.0= 7.21 + log10 ([A-] / [HA] ) = -0.21 = log10([ A-] / [HA] ) ([A-] / [HA] ) = 10^-0.21= 0.616 Method 1: Final volume x concentration x nw = mass 0.616 [HA] + [A-] = 0.1 m : 1.616[HA]= 0.1 m [HA] 0.1m / 0.616= 0.062 M [A-] = 0.616 m x 0.062m = 0.038 M Now , [HA]= 50 / 1000 L x 0.062m x 136g/ mol = 0.4216 g [A-] = 50/1000 L x 0.038 m x 174 g/ mol = 0.330 g Method 2 Equation: C1V1= C2V2 [A-1] : V1= 0.062m x 50 mL = 3.1 mL [ HA] : V1 = 0.038 m x 50 mL = 1.9 mL Method 3

mass = volume x concentration x nw [A-] = 50 / 1000 mL x 0.1m x 136 g/ Mol = 0.68 g [HA] = 50/1000 mL x 0.1m x 174g/mol = 0.87 g Table 2: Ph measurements of buffers Buffer Measured Ph Method 1 (WA+WB) 6.49 Method 2 (WB +SA) 6.58 Method 3 (WA+SB) 6.6 Average 6.56 Table 3 Buffer Capacity Calculated pKa was measured by subtracting the literature pKa minus the pH given then subtract the average pH in Table 2. Literature Pka - ( pH given - average pH ) 7.21 -( 7.0- 6.56 ) =6.77 Buffer Starting Ph Measured Ph after addition Calculated Ph Buffer + 0.5 mL HCI 6.58 6.51 6.77 Buffer + 0.5 mL NaOH 6.6 6.9 6.77 Water + 50 μL HCI 7.58 3.02 6.77 Water + 50 μL NaOH 7.58 10.86 6.77 Table 4: Buffer Capacity Calculation

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The initial calculation was determined by using the initial values in Method 1. [HA] = ( 50/ 1000 L x 0.062 x 1000) = 3.1 [A-] = (50 /1000 L x 0.038 m x 1000 ) = 1.9 The change in mmol was determined by the equation below. 10^-6 x 500 microliters x 1 mol/ L x 1000 mmol = 0.5 mmol. The end mmol was determined by subtracting the change from the initial values. The above equations were used to determine the values of buffer solutions. However the same process was used for unbuffered calculations, but instead of using 500 microliters, use 50. %0 microliters were added to the capacity experiment. HA Weak Acid ( buffered) A- Weak base ( buffered ) HA Weak Acid ( unbuffered) A- Weak base ( unbuffered ) Initial mmol 3.1 1.9 3.05 1.85 Change in mmol 0.5 0.5 0.05 0.05 End mmol 2.6 1.4 2.55 1.35 Discussion The purpose of the buffer lab was to prepare and measure the pH of three buffer types. The calculations were found using the henderson-hasselbalch equation. Stoichiometry was used for the solid potassium phosphate dibasic and monobasic, however amounts were added to a liquid volume since it is solid. The liquid ingredients HCI and NaOH were massed using the calculation form of C1V1= C2V2. During the lab, grams and milliliters were used to calculate each buffer. Experimental errors could have occurred when calculating or rounding certain digits

before the final calculations. However, there was a minor error in the process when calculating the mass amounts at the beginning of the lab. When calculating the buffer capacity the henderson-hasselbalch equation was used again to measure the three acids and the two acids with the pH closest to each other were chosen. During the experiment HCI was added to one of the chosen and the other chosen acid was added to NaOH then measured the pH for each new buffer. Buffer capacity calculations The pK2 was calculated by using the assigned pH of 7.0 and subtracting the average pH, then the literature pKa of 7.21 was subtracted. Example, ( 7.21- (7.0- 6.56)). The pH values obtained were all less than the assigned pH given. This means that my prediction at the beginning of the lab was satisfied. However, there might have been errors when conducting my experiment such as errors with liquid transfer, solid massing errors or pH meter reading errors. The experiment was designed and established to understand acid and base concentrations and the usage of henderson-hasselbalch equation. The provided beakers were measured to be 50 mL. The solution volume was also 50 mL. This can cause accidents of spilling or splashing when stirring or transferring the products between the cylinder and the beaker. It would be beneficial to have larger beakers to hold the buffer solution. Questions: 1. Your goal is to make 100 ml of 0.1 M buffer at a pH of 4.7 using method #2. You are given sodium acetate, HCl (1 M), The sodium salt of bicarbonate - NaHCO3 (salt of the weak acid), and Sodium carbonate - Na2CO3 (its conjugate base). Determine which of these four you will need and then show the calculations including the pKa value of the chosen buffer.

The concentration of acid and salt should be the same. The concentration of sodium acetate = 0.05 acetic acid. The volume given is 100 mL, Moles = 0.005 moles. Take 0.005 mL of the 50 mL sodium acetate and react it with 50 mL of 0.05M HCl. 2. 2. You are given sodium acetate, HCl (1 M), NaHCO3, and Na2CO3. Determine which of these four you will need and then show the calculations(include the pKa of the chosen buffer) to make a buffer of pH 9.9 using method #2. Assume you are making 100 ml of a 0.1 M buffer. Equation used: pH=pKa+log ([salt])/([acid]) 9.9=10.1+log ([Na2CO3])/([NaHCO3]) —> ([Na2CO3])/([NaHCO3]) =0.63 [Na2CO3]+{NaHCO3]=0.1M 0.63[NaHCO3]+[NaHCO3]=0.1—-> [NaHCO3]= 0.0613M Moles=Molarity x Volume= 0.0613 x 100/1000=0.00613 moles [Na2CO3]=0.1-0.613=0.938 M or moles=0.0938 3. 3. By consulting a buffer table of pKa values, suggest a buffer system other than phosphate or carbonate and calculate the raGo of the conjugate base/acid concentrations to obtain a buffer at pH 6.6. pH = pKa + log [base]/[acid] —> 6.6 = 6.49 + log [base]/[acid] → log [base]/ [acid] = 6.6-6.49 [ A- ]/[0.5 M ] =1.29 -> 0.65 M

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