A beaker with 1.80×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.50 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

The pH will change 0.39

Explanation:

Step 1: Data given

Volume of acetic acid : 180 mL

pH = 5.00

The total molarity of acid and conjugate base in this buffer is 0.100 M.

We add 8.50 mL (= 0.0085 L) of a 0.470 M HCl

pKa of acetic acid = 4.74

Step 2: Calculate initial concentrations

x = concentration acid

y = concentration conjugate base

x + y = 0.100

pH = pKa = log ([y]/[x])

5.00 = 4.74 + log y/x

 0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

x + 1.82 x = 0.100

2.82 x = 0.100

x = 0.0355 = concentration of acid

y = 0.0645 = concentration of conjugate base

Step 3: Calculate moles

Moles = molarity * volume

moles acid = 0.180 L * 0.0355 M= 0.00639 moles

moles conjugate base = 0.0645 M * 0.180 L=0.01161 moles

moles HCl = 8.50 * 10^-3 L * 0.470 M=0.003995 moles

A- + H+ = HA

moles conjugate base = 0.01161- 0.003995=0.007615  moles

moles acid = 0.00639 + 0.003995=0.010385moles

Step 4: Calculate total volume

total volume = 180 + 8.50 = 188.5 mL = 0.1885 L

Step 5: Calculate concentration

concentration acid = 0.010385 moles/ 0.1885 L =0.0551 M

concentration conjugate base = 0.007615/ 0.1885 =0.0404 M

Step 6: Calculate pH

pH = 4.74 + log 0.0404/0.0551=4.61

Step 7: Calculate pH change

change pH = 5.00 - 4.61=0.39

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.