Part A beaker with 175 ml of an acetic aoid buter with a pH of 5,000 is itng on

Question

Part A beaker with 175 ml of an acetic aoid buter with a pH of 5,000 is itng on a benchtop The total molarity of acid and conugate base in this buter in 0 00 A student adds 5.30 ml of a 0.300 MHCI solution to the beaker How much wll the pht change? The pK, of acec od n 4.740 Express your anw war nunartally to two decimal place. Use minus (-) the pH has decom View Available Hint) conjugate base to counteract the offlects of acid or base addition on ph on CHa COO added in a salt form (e.g, sodium acetabe NaCH,COO) the addtion oH' or OH lons Thun, for conjugate base s desorbed by the Hendrson Hasselich

Explanation / Answer

Let [HA] be acetic acid and [A-] is the conjugate base.

From Henderson Hasselbalch equation, pH =pKa+ log { [A-]/[HA]}   (1)

Given pH= 5,pKa =4.74, hence from Eq.15= 4.74+ log {[A-]/[HA]}

[A-]/[HA]= 1.82, [A-]= 1.82[HA] (2)

Given [A-]+[HA] =0.1 from Eq.2, [HA]+1.82[HA]= 0.1, [HA]= 0.0354M, [A-]= 1.82*0.0354=0.0646M

Now volume of buffer= 175ml, since 1000ml =1L, volume of buffer= 175/1000=0.175L

Moles : [HA]=molarity* volume in Liters = 0.0354*0.175 =0.0062, moles of [A-]=0.0646*0.175=.0113

Moles of HCl added= 0.3*5.3/1000=0.00159, the HCl added reacts with [A-] and forms HA.

HCl+A- -----àHA, molar ratio of HCl:A- = 1:1 ( Theoretical), actual =0.00159:0.0113 =1:7.1. So excess is A- . All the HCl added reacts.

So this will add to additional moles of HA forned. Total moles of HA formed= 0.0062+0.00159=0.00779, moles of A- remaining = 0.0113-0.00159=0.00971

Volume of solution after mixing = 175+5.3=180.3 ml= 180.3/1000L=0.1803L

Concentration= moles/Volume, [HA]=0.00779/0.1803M and [A-]= 0.00971/0.1803

Now from Eq.1, new pH= 4.74+log (0.00971/0.00779)=4.84

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