Solution C is a 1.00 L buffer solution that is 1.420 M in acetic acid and 0.67 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH change of this solution upon addition of 0.10100 mol of HCl? Enter a negative number to 3 decimal places.

The correct answer is 0.10.

Explanation:

Based on the given question, in a buffer solution of 1 liter, the molarity of acetic acid is 1.420 M and the molarity of sodium acetate is 0.67. The pKa value of acetic acid given is 4.74, now the pH of buffer is,  

pH of buffer = pKa + log ([CH3COONa]/[CH3COOH])

= 4.74 + log (0.67/1.420)

= 4.74 + (-0.326)

= 4.41  

Now 0.10100 mol of HCl is added, the HCl reacts with sodium acetate to give,  

CH3COONa + HCl = CH3COOH + NaCl

Now the concentration of CH3COONa becomes = 0.67-0.101 = 0.57 M, and the new concentration of CH3COOH becomes = 1.420 + 0.101 = 1.52 M

Now the new pH will be,  

= pKa + log (0.57/1.52)

= 4.74 + (-0.426)

= 4.31

The pH change is 4.41-4.31 = 0.10  

answer: the primary difference between a hypothesis and a theory is that b hypothesis is falsifiable, but a theory has been proven to be correct. theory is an explanation of something that is happening but hypothesis is a guess of something that can happen

explanation:

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