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

Question

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

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1. a. pH = 12.711

b. pH = 1.769

### Further explanationn

pH is the degree of acidity of a solution that depends on the concentration of Hu207a ions. The greater the value the more acidic the solution and the smaller the pH.n

pH = – log [Hu207a]n

So that the two quantities between pH and [Hu207a] are inversely proportional because they are associated with negative values.n

A solution whose value is different by n has a difference in the concentration of Hu207a ion of 10u207f.n

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

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 OHu207b / [OHu207b] by using the formula pH = 14-pOHn

[OH u207b] = b. Mb wheren

b = number of OHu207bn

Mb = strong base concentrationn

2. If the rest of the reaction results obtained the remaining strong acid HCl, the pH is sought from the concentration of Hu207a / [Hu207a] by using the formulan

[Hu207a] = a. Mn

a = valence of acid / amount of Hu207a releasedn

M = acid concentration.n

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

n

n

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

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

b) 24.0mL of 0.440 M NaOH (aq)n

n

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

mol NaOH = 44.0 ml x 0.340 M = 14.96 mmoln

The remaining moles: 14.96 – 11.56 = 3.4 mmoln

[OH-] = mmol / total volumen

[OH-] = 3.4 / 44 + 34n

[OH-] = 0.0513n

pOH = – log 5.13.10-2n

pOH = 1,289n

pH = 14-1.289 = 12.711

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

mol NaOH = 24.0mL x 0.440 M = 10.56 mmoln

The remaining moles: 11.56 -11.56 = 1 mmoln

[H+] = mmol / total volumen

[H+] = 1 / 24 + 34n

[H+] =0.017

pH = – log 0.017

pH = 1.769

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Keywords : pH, acid, base, HCl,NaOH

2. According to the reaction equation:

HCl(aq) + NaOH(aq)u2192 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:

u2234 NaOH remaining = 0.01326 – 0.01156

= 0.0017 mol

when the total volume = 0.039 + 0.034

= 0.073 L

u2234[OH] = moles / total volume

= 0.0017mol / 0.073 L

= 0.0233 M

u2234 POH = -u33d2[OH-]

= -u33d20.0233

= 1.63

u2234 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
u2234[OH-] = 0.0056 / 0.078

= 0.0718 M

u2234 POH = -u33d2[OH-]

= -u33d20.0718

= 1.14

u2234PH = 14 – POH

= 14 -POH

= 14 – 1.14

= 12.86