50 ml of 1n ch3cooh mixed with 10 of 1 n ch3coona solution wil have an approximate ph (pka of ch3cooh=4.7 and log2=0.3)
a) 4
b) 5
c) 3
d) 7

The solution of a weak acid and its salt with strong base will make a buffer solution

Buffer solution is the solution which resists change in pH on addition of small amount of acid or base

the pH of buffer solution can be calculated using Hendersen Hassalbalch's equation

the equation is

given

pKa = 4.7

[salt] = moles of salt / total volume

moles of salt = molarity X volume = 1 X 10 = 10mmol

[salt]= 10/60 = 0.167

[acid] = moles of acid / total volume

moles of acid = molarity X volume = 1 X 50 = 50mmol

[acid]=50/60 = 0.833

pH = 4.7 + log [0.833/00167] = 4.7-0.698 =4

pH will be 4

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol