To prepare a buffer solution of PH 8.26, amount of (NH4)2SO4 to be add

1.	To prepare a buffer solution of PH 8.26, amount of (NH4)2SO4 to be added into 500ml of 0.01M NH4OH solution if pkb(NH4+) is 9.26
Answer» Answer:- $1.99510^-^4g$ Solution:- The pH of a buffer solution is calculated by using the Handerson equation: $pH=pKa+log(\frac{base}{acid})$ For the given buffer, the weak base is ammonium hydroxide and the acid is ammonium sulfate as it is a salt of weak base(ammonium hydroxide) and strong acid(sulfuric acid). moles of ammonium hydroxide can be calculated from its given volume and MOLARITY as: $500mL(\frac{1L}{1000mL})(\frac{0.01mol}{L})$ = 0.005 mol pKa is calculated from given pKb using the formula: pKa = 14 - pKb pKa = 14 - 9.26 pKa = 4.74 pH is given as 8.26, LET's plug in the given values in handerson equation and CALCULATE the moles of acid that is ammonium sulfate. $8.26=4.74+log(\frac{0.005}{n})$ $8.26-4.74=log(\frac{0.005}{n})$ $3.52=log(\frac{0.005}{n})$ On taking antilog: $10^3^.^5^2=log(\frac{0.005}{n})$ $3311.31=(\frac{0.005}{n})$ $n=\frac{0.005}{3311.31}$ $n=1.5110^-^6$ Molar MASS of $(NH_4)_2SO_4$ is 132.14 GRAM per mol. Let's multiply the moles by molar mass to get the mass in grams. $1.5110^-^6mol(\frac{132.14g}{mol})$ = $1.99510^-^4g$ $1.99510^-^4g$ So, of ammonium sulfate are required to prepare the buffer.*

To prepare a buffer solution of PH 8.26, amount of (NH4)2SO4 to be added into 500ml of 0.01M NH4OH solution if pkb(NH4+) is 9.26

Answer»

Answer:- $1.995*10^-^4g$

Solution:- The pH of a buffer solution is calculated by using the Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

For the given buffer, the weak base is ammonium hydroxide and the acid is ammonium sulfate as it is a salt of weak base(ammonium hydroxide) and strong acid(sulfuric acid).

moles of ammonium hydroxide can be calculated from its given volume and MOLARITY as:

$500mL(\frac{1L}{1000mL})(\frac{0.01mol}{L})$