Question

Calculate the ph of the resulting solution if 34.0 ml of 0.340 m hcl(aq is added to

Answer 1

According to the reaction equation:

HCl(aq) + NaOH(aq) →  NaCl(aq) +  H2O(l)

a) Part 1):

first, we need to get moles HCl = molarity * volume

                                                      = 0.34m * 0.034

                                                      = 0.01156 mol

then, we need moles of NaOH = molarity * volume

                                                    =  0.34 * 0.039

                                                     = 0.01326 mol

when NaOH excess:

∴ NaOH remaining = 0.01326 - 0.01156

                                 = 0.0017 mol 

when the total volume = 0.039 + 0.034

                                       =  0.073 L

∴[OH] = moles / total volume 

           = 0.0017mol / 0.073 L

           =  0.0233 M

∴ POH = -㏒[OH-]

             = -㏒0.0233

             = 1.63 

∴ PH = 14- 1.63

         = 12.37 

b) part 2:

as we got moles HCl = molarity * volume 

                                   = 0.34 * 0.034

                                  = 0.01156 mol 

then moles NaOH = molarity * volume 

                               = 0.39 * 0.044

                               = 0.01716 mol

NaOH remaining = 0.01716 - 0.01156

                             = 0.0056 mol

when the total volume = 0.034 + 0.044

                                       = 0.078 L
∴[OH-] = 0.0056 / 0.078

            = 0.0718 M

∴ POH = -㏒[OH-]

            = -㏒0.0718

            = 1.14

∴PH = 14 - POH

        = 14 -POH

       = 14 - 1.14

       = 12.86

Answer 2

a. pH = 12.711

b. pH = 1.769

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 remainder of the reaction product is obtained the remaining strong base of NaOH then the pH is sought from the concentration of OH⁻ / [OH⁻] by using the formula pH = 14-pOH

[OH ⁻] = b. Mb where

b = number of OH⁻

Mb = strong base concentration

2. If the rest of the reaction results obtained the remaining strong acid HCl, the pH is sought from the concentration of H⁺ / [H⁺] by using the formula

[H⁺] = a. M

a = valence of acid / amount of H⁺ released

M = acid concentration.

3. if both of them run out of reaction, then pH = 7

We complete the answer choices

Calculate the pH of the resulting solution if 34.0 mL of 0.340 M HCL (aq) is added to

a) 44.0mL of 0.340 M NaOH (aq).

b) 24.0mL of 0.440 M NaOH (aq)

a. mole HCl = 34.0mL x 0.340 M = 11.56 mmol

mol NaOH = 44.0 ml x 0.340 M = 14.96 mmol

The remaining moles: 14.96 - 11.56 = 3.4 mmol

[OH-] = mmol / total volume

[OH-] = 3.4 / 44 + 34

[OH-] = 0.0513

pOH = - log 5.13.10-2

pOH = 1,289

pH = 14-1.289 = 12.711

b. mole HCl = 34.0mL x 0.340 M = 11.56 mmol

mol NaOH = 24.0mL x 0.440 M = 10.56 mmol

The remaining moles: 11.56 -11.56 = 1 mmol

[H+] = mmol / total volume

[H+] = 1 / 24 + 34

[H+] =0.017

pH = - log 0.017

pH = 1.769

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Keywords : pH, acid, base, HCl,NaOH