Chemistry, 27.07.2019 23:00 gomez36495983

Calculate the ph of the resulting solution if 27.0 ml of 0.270 m hcl(aq is added to 37.0 ml of 0.270 m naoh(aq

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Answer from: ilovefood2064

The reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.
Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol
number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol
HCl reacts with NaOH in 1:1 molar ratio
Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol
total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L
Since there's excess OH⁻ ions, we can calculate pOH value first
pOH = - log [OH⁻]
[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L
pOH = -log(0.042 M)
pOH = 1.37
by knowing pOH we can calculate pH using the following equation;
pH + pOH = 14
pH = 14 - 1.37
pH = 12.63

Answer from: DisneyGirl11

pH of the resulting solution : 12.625

Further explanation

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

A solution whose value is different by n has a difference in the concentration of H⁺ ion of 10ⁿ.

The pH value of a reaction between strong acid HCl and strong base NaOH can be estimated from the rest of the reaction product

1. If the remaining reaction product is a strong base (NaOH) then the pH is obtained from the concentration of OH⁻ / [OH⁻] by using the formula pH = 14-pOH

[OH ⁻] = b. Mb

b = number of OH⁻

Mb = strong base concentration

2. If the remainder of the reaction product is strong acid HCl, the pH is obtained from the concentration of H⁺ / [H⁺] using the formula

[H⁺] = a. M

a = valence of acid / amount of H⁺ released

M = acid concentration.

3. if both (acids and bases) run out of reaction, then pH = 7

27.0 ml of 0.270 m HCl (aq) is added to 37.0 ml of 0.270 m NaOH (aq)

a. mole HCl = 27.0mL x 0.27 M = 7.29 mmol

mol NaOH = 37 ml x 0.27 M = 9.99 mmol

The remaining moles: 9.99 - 7.29 = 2.7 mmol ⇒there is remaining NaOH

so pH from [OH-]

$\displaystyle [OH-]=\frac{mmol}{total~volume}\\\\=\frac{2.7}{27+37}\\\\ =0.0422\\\\pOH=-log~0.0422\\\\pOH=1.375\\\\pH=14-1.375\\\\pH=\boxed{\bold{12.625}}$

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Calculate the ph of the resulting solution if 27.0 ml of 0.270 m hcl(aq is added to 37.0 ml of 0.270...

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