You have a 0.5 l of a 1.0 m buffer solution with pka = 5.14 and current ph = 5.23. calculate the new ph when 1.50 ml of 5.00 m hcl is added. be mindful of units. (how can you tell if your answer makes sense? ) ph = pka + 109 u base 5.23 - 5.14+ log ( )

pH = 5,22

Explanation:

A buffer consist in a solution with both conjugate acid (HA) and conjugate base (A⁻). To know the concentration of both A⁻ and HA you use Henderson-Hasselbalch equation:

pH = pka + log₁₀

In the problem:

5,23 = 5,14 + log₁₀

1,23 = (1)

Also, you know buffer concentration is 1,0 M:

1,0 M = A⁻ + HA (2)

Replacing (2) in (1):

HA = 0,45 M, thus:

A⁻ = 0,55 M.

The addition of 1.50 mL of 5.00 M HCl represents a concentration of:

1,50x10⁻³ L× ÷ 0,5 L = 0,015 M of HCl

The buffer equilibrium is:

HA ⇄ A⁻ + H⁺ ka =

The concentrations in equilibrium are:

[HA] = 0,45 M + x

[A⁻] = 0,55 M - x

[H⁺] = 0,015 M -x

Because the equilibrium is displaced to the left.

The equation of equilibrium is:

=

You will obtain:

x² - 0,565x + 8,24674x10⁻³ = 0

Solving:

x = 0,5500061 ⇒ No physical sense

x = 0,01499391

Thus, [H⁺] = 0,015 - 0,01499391 = 6,09x10⁻⁶

As pH = -log₁₀ [H⁺]

pH = 5,22

I hope it helps!

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

fr8t7iot7o87oglgt87ot79p0[90